Embed Size (px)
Citation preview
Study
Steps
.in
Parabola
1. CONIC SECTIONThe locus of a point, which moves so that its distance from a fixed point is always in a constant ratioto its distance from a fixed straight line, not passing through the fixed point is called a conic section.
The fixed point is called the focus.
The fixed straight line is called the directrix.
The constant ratio is called the eccentricity and is denoted by e.
When the eccentricity is unity i.e., e = 1, the conic is called a parabola ; when e < 1 the conicis called an ellipse and when e > 1, the conic is called a hyperbola.
The straight line passing through the focus and perpendicular to the directrix is called the axisof the parabola
A point of intersection of a conic with its axis is called vertex.
2. STANDARD EQUATION OFA PARABOLALet S be the focus, ZM the directrix and P the moving point. Draw SZ perpendicular from S on thedirectrix. Then SZ is the axis of the parabola. Now the middle point of SZ, say A, will lie on the locusof P, i.e., AS = AZ. Take A as the origin, the x-axis along AS, and the y-axis along the perpendicularto AS at A, as in the figure.
Z A S
M YPNX
Let AS = a, so that ZA also a. Let (x, y) be the coordinates of the moving point P. ThenMP = ZN = ZA + AN = a + x. But by definition MP = PS MP2 = PS2
So that, (a + x)2 = (x – a)2 + y2.Hence, the equation of parabola is y2 = 4ax.
2.1 LATUS RECTUMThe chord of a parabola through the focus and perpendicular to the axis is called the latus rectum.
In the figure LSL is the latus rectum.
Also LSL = 2 a.a4 = 4a = double ordinate through the focus S.
Note :
Any chord of the parabola y2 = 4ax perpendicular to the axis of the parabola is called double
Theory Notes - Parabola
Page 1 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
ordinate.
Two parabolas are said to be equal when their latus recta are equal.
2.2 Four Common forms of a Parabola :y2 = 4ax y2 = –4ax x2 = 4ay x2 = –4ay
Vertex : (0 , 0) (0, 0) (0, 0) (0, 0) Focus : (a, 0) (– a, 0) (0, a) (0, – a) Equation of the Directrix x = – a x = a y = – a y =a Equation of the axis y = 0 y = 0 x = 0 x = 0 Tangent at the vertex x = 0 x = 0 y = 0 y = 0
Illustration 1:Find the vertex, axis, directrix, focus, latus rectum and the tangent at vertex for the parabola9y2 – 16x – 12y – 57 = 0
Solution :
The given equation can be rewritten as2
3
2y
=
9
16
16
61x
which is of the form Y2 = 4AX
Hence the vertex is
3
2,
16
61
The axis is y –3
2 = 0 y =
3
2
The directrix is X + A = 0 x +16
61 =
9
4 x =
144
485
Also
3
2,
144
485 is the focus.
Length of the latus rectum = 4A =9
16
The tangent at the vertex is X = 0 x = –16
61.
Illustration 2:The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the equation of theparabola and the points where it meets the axes.
Solution :Focus of the parabola is the mid-point of the latus rectum. S is (7, 4). Also axis of the parabola is perpendicular to the latus rectum and passes throughthe focus. Its equation is
y – 4 =35
0
(x – 7) y = 4
Length of the latus rectum = (5 – 3) = 2Hence the vertex of the parabola is at a distance 2/4 = 0.5 from the focus. We have two parabola,
Page 2 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
one concave rightwards and the other concave leftwards. The vertex of the first parabola is(6.5, 4) and its equation is (y – 4)2 = 2(x – 6.5) and it meets the x-axis at (14.5, 0). The equation ofthe second parabola is (y – 4)2 = – 2 (x – 7.5). It meets the x-axis at (– 0.5, 0) and the y-axis at (0.4, ± 5 ).
Drill Exercise - 1
1. Find the vertex, axis, focus, directrix, latusrectum of the following parabolas. Also draw their roughsketches(i) 4y2 + 12x – 20y + 67 = 0 (ii) y = x2 – 2x + 3
2. Find the equation of the parabola with vertex (2, –3) and focus (0, 5).
3. Find the equation of the parabola whose focus is (1, – 1) and whose vertex is (2, 1). Also find its axisand latus rectum.
4. Find the equation of the parabola whose focus is the point (0, 0) and whose directrix is the straight line3x – 4y + 2 = 0.
5. Find the equation of the parabola whose latusrectum is 4 units, axis is the line 3x + 4y – 4 = 0 and thetangent at the vertex is the line 4x – 3y + 7 = 0.
2.3 Parametric CoordinatesAny point on the parabola y2 = 4ax is (at2 , 2at) and we refer to it as the point ‘t’. Here, t is aparameter, i.e., it varies from point to point.
Illustration 3:Show that the locus of a point that divides a chord of slope 2 of the parabola y2 = 4x internally in theratio 1 : 2 is a parabola. Find the vertex of this parabola.
Solution :Let the two points on the given parabola be (t
12, 2t
1) and (t
22 , 2t
2). Slope of the line joining these point
is = 21
22
12
tt
t2t2
=21 tt
2
t1 + t
2 = 1. Hence the two points become (t
12 , 2t
1) and ((1 – t
1)2 , 2(1 – t
1)). Let (h, k) be the
point which divides these points in the ration 1 : 2.
h =3
t2)t1( 21
21
=3
t3t21 211
.......(1)
k =3
t4)t1(2 11 =
3
t22 1........(2)
Eliminating t1 from (1) and (2), we find that 4h = 9k2 – 16k + 8
Hence locus of (h,k) is (y – 8/9)2 =9
4
9
2x . This is a parabola with vertex
9
8,
9
2.
2.4 Focal Chord :
Page 3 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Any chord to y2 = 4ax which passes through the focus is called a focal chord of the parabolay2 = 4ax .Let y2 = 4ax be the equation of a parabola and (at2, 2at) a point P on it. Suppose the coordinates ofthe other extremity Q of the focal chord through P are (at
12, 2at
1).
Then, PS and SQ, where S is the focus (a, 0) have the same slopes.
aat
0at22
= aat
0at221
tt
12 – t = t
1t2 – t
1 (tt
1 + 1)(t
1 – t) = 0
Hence t1 = – 1/t, i.e. the point Q is (a/t2, – 2a/t).
The extremities of a focal chord of the parabola y2 = 4ax may be taken as the points t and – 1/t.
Illustration 4:Through the vertex O of a parabola y2 = 4x chords OP and OQ are drawn at right angles to oneanother. Show that for all position of P, PQ cuts the axis of the parabola at a fixed point. Also find thelocus of the middle point of PQ.
Solution :The given parabola is y2 = 4x ........(1)Let P (t
12, 2t
1), Q (t
22, 2t
2)
Slope of OP = 21
1
t
t2 =
1t
2
Since OP OQ,21tt
4 = – 1 or t
1t2 = – 4 ........(2)
The equation of PQ is y – 2t1 = 2
221
21
tt
)tt(2
(x – t12)
or y (t1 + t
2) = 2 (x + t
1t2) y (t
1 + t
2) = 2 (x – 4) [from (2)]
This line cuts the x-axis (the axis of the parabola) at (4, 0) which is a fixed point for all positions of P.
Let R() be the middle point of PQ. Then =2
tt 22
21
........(3)
and = t1 + t
2
From (4), 2 = t12 + t
22 + 2t
1 t
2 = 2 – 8 [From (2) and (3)]
Hence Locus of R (, ) is y2 = 2x – 8
2.5 Focal Distance of Any Point :The focal distance of any point P on the parabola y2 = 4ax is the
Page 4 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
distance between the point P and the focus S, i.e. PS.
Thus the focal distance = PS = PM = ZN = ZA + AN = a + x
Illustration 5:The focal distance of a point on the parabola y2 = 12x is 4. find the abscissa of this point.
Solution :On comparing y2 = 12x with the standard form y2 = 4ax we get a = 3If the given point on y2 = 12x is (x, y) then its focal distance is x + 3. x + 3 = 4 x = 1.Hence the abscissa of the given point is 1.
Drill Exercise - 2
1. If the focal distance of a point on the parabola y2 = 4x is 10 units, then find its coordinates.
2. If the focal distance of a point on the parabola x2 = 16y is 3 units, then find its coordinates.
3. PQ is a variable focal chord of the parabola y2 = 4ax whose vertex is A. Prove that the locus of thecentroid of triangle APQ is a parabola whose latus rectum is 12/9 a.
4. Show that the semilatusrectum of the parabola y2 = 4ax is the harmonic mean between the segmentsof any focal chord of the parabola.
5. Show that the focal chord of parabola y2 = 4ax makes an angle with the x-axis is of length 4acosec2 .
2.6 Position of a Point Relative to the Parabola :Consider the parabola : y2 = 4ax. If (x
1, y
1) is a given point and y
12 – 4ax
1 = 0, then the point lies on
the parabola. But when y12 – 4ax
1 0. We draw the ordinate PM meeting the curve in L. Then P will
lie outside the parabola if PM > LM, i.e., PM2 – LM2 > 0
Now, PM2 = y1
2 and LM2 = 4ax1 by virtue of the parabola. Substituting these values in equation of
parabola, the condition for P to lie outside the parabola becomes y12 – 4ax
1 > 0.
Similarly, the condition for P to lie inside the parabola is y12 – 4ax
1 < 0.
Page 5 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Illustration 6:For what values of ‘’ the point P(, ) lies inside, on or outside the parabola (y – 2)2
= 4(x – 3).
Solution :Given equation can be written as y2 – 4y – 4x + 16 = 0Point P(, ) lies inside parabola if 2 – 8 + 16 < 0 ( – 4)2 < 0 no such exist.Point P () lies on parabola if ( – 4)2 = 0 = 4Point P() lies outside parabola if ( – 4)2 > 0 R – {4}
3. THE GENERAL EQUATION OFA PARABOLAWe shall now obtain the equation of a parabola when the focus is any point and the directrix is anyline. Let (h, k) be the focus S and lx + my + n = 0, the equation of the directrix ZM of a parabola. Let(x, y) be the coordinates of any point P on the parabola. Then the relation,PS = distance of P from ZM, gives (x – h)2 + (y – k)2 = (lx + my + n)2/(l2 + m2) (mx – ly)2 + 2gx + 2fy + d = 0
This is the general equation of a parabola. It is clear that second-degree terms in the equation of aparabola form a perfect square. The converse is also true, i.e. if in an equation of the second degree,the second degree terms form a perfect square then the equation represents a parabola, unless itrepresents two parabola, unless it represents two parallel straight lines.
Note : The general equation of second degree i.e. ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a parabola
if 0 and h2 = ab, =cfg
fbh
gha
.
Special case :Let the vertex be () and the axis be parallel to the x-axis. Then the equation of parabola is givenby (y – b)2 = 4a (x – ) which is equivalent to x = Ay2 + By + CIf three points are given we can find A, B and C.Similarly, when the axis is parallel to the y-axis, the equation of parabola is y = Ax2 + Bx + C
Illustration 7:Find the equation of the parabola whose focus is (3, – 4) and directrix x – y + 5 = 0.
Solution :Let P(x, y) be any point on the parabola. Then
22 )4y()3x( =11
|5yx|
Page 6 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
(x – 3)2 + (y + 4)2 =2
)5yx( 2
x2 + y2 + 2xy – 22x + 26y + 25 = 0 (x + y)2 = 22x – 26y – 25
Drill Exercise - 3
1. Find the name of the conic represented by the equation x2 + y2 – 2xy + 20x + 10 = 0.
2. The point (–2m, m + 1) is an interior point of the smaller region bounded by the circle x2 + y2 = 4 andthe parabola y2 = 4x. Then find the interval in which m lies.
3. The point (a, –1) is extension to both parabolas y2 = | x | , then find the set of all possible real valuesof a.
4. Find the set of values of in the interval [/2, 3/2], for which the point (sin , cos ) does not lieoutside the parabola 2y2 + x – 2 = 0.
5. If (a2, a – 2) be a point interior to the region of the parabola y2 = 2x bounded by the chord joining thepoints (2, 2) and (8, –4), then find the set of all possible real values of a.
4. TANGENT DRAWNATA POINT LYING ON A GIVEN PARABOLA :(i) If P (x
1, y
1) be a point on the parabola y2 = 4ax, then the equation of the tangent at P is
yy1 = 2a (x + x
1)
i.e T(x,y) = 0
(ii) If P(at2, 2at) be any point on the parabola y2 = 4ax, then , slope of the tangent at
P =at2
a2 =
t
1
y
a2
dx
dy
and hence its equation is
y – 2at =
t
1 (x – at2)
i.e. yt = x + at2 .........(1)
If we substitute m fort
1, in equation (1), we have the following result. The equation
y = mx +
m
a........(2)
4.1 Point of Intersection of Tangents at ‘t1’ & ‘t2’Equations of the tangents at the points (at
12, 2at
1) and (at
22 , 2at
2) are (by ARt. 7.6)
Page 7 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
yt1 = x + at
12 .........(4)
and yt2 = x + at
22 .........(5)
respectively.Solving equations (4) and (5) gives the coordinates of the intersection point of these two tangents as{at
1t2, a(t
1 + t
2)}
Illustration 8:Tangents at the extremities of any focal chord intersect at right angles on the directrix.
Solution :Let P(at2, 2at) and P ( 2
1at , 2at1 ) be the end points of a focal chord of the parabola.
Then t.t1 = –1. Equation of the tangent at the point P and the point P are ty = x + at2 and
t1y = x + at
12 respectively.
Let these tangents intersect at a point (h, k). Then h = at1t and k = a(t + t
1). Now here
1t
1.
t
1 = – 1 i.e., the tangents are perpendicular and also h = –a. Hence the locus of the point (h, k) is
x = –a which is the equation of the directrix.
Note: (i) If the tangents at t1 & t
2 are at right angles then t
1t2 = –1.
(ii) If the chord joining t1, t
2 subtends a right angle at the vertex then t
1t2 = – 4
Drill Exercise - 4
1. Prove that the point of intersection of tangents at two points P (at1
2, 2at1) and Q (at
22, 2at
2) on the
parabola y2 = 4ax is the point R (at1t2, a (t
1+ t
2)).
2. The curves y2 = 4x, x2 = 4y cut in two points. Find the angles between the tangents at each point ofintersection.
3. If the tangents at the points P and Q on a parabola meet in T, prove that ST2 = SP. SQ i.e., ST is thegeometric mean of the focal distances of P and Q.
4. If the tangents to the parabola y2 = 4ax at the points P and Q intersects at T, prove that TP and TQsubtend equal angles at the focus.
5. Show that the locus of the point of intersection of perpendicular tangents to the parabola is its directrix.
4.2 Intersection of a Line and a parabolaLet the parabola be y2 = 4ax ......(1)and the given line be y = mx + c ......(2)Eliminating y from (1) and (2), then
(mx + c)2 = 4axor m2x2 + 2x (m c– 2a) + c2 = 0 ......(3)This equation is quadratic in x, gives two values of x. Which shows that every straight line will cut theparabola in two points may be real, coincident or imaginary according as Discriminant of(3) > , = , < 0i.e. 4(mc – 2a)2 – 4m2c2 > , = , < 0or 4a2 – 4amc > , = , < 0
Page 8 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
or a > , = , < mc ......(4)
Note : If m = 0 then equation (3) gives – 4ax + c2 = 0 or x =a4
c2
which gives only one value of x and so every
line parallel to x-axis cuts the parabola only in one real point.
4.3 Condition of TangencyIf the line (2) touches the parabola (1), then equation (3) has equal rootsDiscriminant of (3) = 0 4(mc – 2a)2 – 4m2c2 = 0 – 4amc + 4a2 = 0
c =m
a, m 0 ......(5)
so, the line y = mx + c touches the parabola y2= 4ax if c =m
a (which is condition of tangency).
Substituting the value of c from (5) in (2) then
y = mx +m
a, m 0 ......(6)
Hence the line y = mx +m
a will always be a tangent to the parabola y2 = 4ax.
4.4 The point of ContactThe point of contact of the tangents at ‘t’ is (at2 , 2at). In terms of slope ‘m’ of the tangent the point
of contact is
m
a2,
m
a2 , (m 0).
Illustration 9:Find the length of the chord of the parabola y2 = 4ax, whose equation is y = mx + c.
Solution :The abscissa of the points common to the straight line y = mx + c and the parabola y2 = 4ax are givenby the equation m2 x2 + (2mc – 4a) x + c2 = 0. If (x
1, y
1) and (x
2, y
2) are the points of intersection,
then (x1 – x
2)2 = (x
1 + x
2)2 –4x
1x
2
42
2
4
2
m
)mca(a16
m
c4
m
)a2mc(4
and (y
1 – y
2) = m(x
1 - x
2)
Hence, the required length
)mca(am1m
4|)xx(|m1)xx()yy( 2
22122
212
21
Illustration 10:Prove that the straight line x + my + n = 0 touches the parabola y2 = 4ax if n = am2.
Solution :The given line is
lx + my + n = 0
Page 9 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
or y = –m
x –
m
n
comparing this line withy = Mx + c .......(1)
M = –m
and c =
m
n
The line (1) will touch the parabola y2 = 4ax, if
c =M
a
or cM = a
or
m
n
m
= a
or n = am2.
Drill Exercise - 5
1. Find the equation of the tangent to the parabola y2 = 12x which is parallel to the line y = x – 3, anddetermine its point of contact.
2. Find the condition that the line ax + by + c = 0 to touch the parabola y2 – 4y – 8x + 32 = 0.
3. Find the equations of the tangents of the parabola y2 = 12 x, which passes through the point (2, 5).
4. Find the equations of the tangents to the parabola y2 = 16x, which are parallel and perpendicularrespectively to the line 2x - y + 5 = 0. Find also the coordinates of their points of contact.
5. Prove that the locus of the point of intersection of tangents to the parabola y2 = 4ax which meet at anangle is (x + a)2 tan2 = y2 – 4ax.
5. EQUATION OF NORMALTO THE PARABOLAIf P(at2, 2at) be any point on the parabola y2 = 4ax, then
slope of the tangent at P =at2
a2 =
t
1.
y
a2
dx
dy
Therefore, slope of the normal at P = – t and its equation isy – 2at = – t(x – at2)
i.e. y = – tx + 2at + at3 .....(1)If we substitute m for – t in equation (1), we have the following result.the equation
y = mx – 2am – am3 .......(2)is a normal real or imaginary can be drawn from any point to a given parabola and the algebraic sumof the ordinates of the feet of these three normals is zero.Let equation of a parabola be
y2 = 4ax ......(3)and that of a normal to it be
Page 10 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
y = mx – 2am – am3 .......(4)If this normal passes through the point (x
1, y
1),we have
y1 = mx
1 – 2am – am3
i.e. am3 + (2a – x1) m + y
1 = 0
This equation gives three values of m, real or imaginary, If m1, m
2 and m
3 be the roots of equation (5),
then we havem
1 + m
2 + m
3 = 0.
Hence, the sum of the ordinates of the feet of these normals = – 2a(m1 + m
2 + m
3) = 0
Note: (i) If normal at the point ‘t1’ meets the parabola again at ‘t
2’ then t
2 = – t
1 –
1t
2
(ii) If the normals at t1& t
2 meet again on the parabola then t
1t2 = 2
(iii) The point of intersection of the normals to the parabola y2 = 4ax at ‘t1’ and ‘t
2’ is
[2a + a(t12 + t
22 + t
1t2), at
1t2 (t
1 + t
2)].
Drill Exercise - 6
1. Prove that the chord of the parabola y2 = 4ax which is normals at the point whose abscissa is 2asubtends a right angle at the vertex.
2. Find the equation of the tangent and normal to the parabola y2 = 16x at the point (1, 4). What lineparallel to this normal touches the parabola ?
3. If the normal at P(18, 12) to the parabola y2 = 8x cuts it again at Q, show that 9PQ = 80 10 .
4. Show that the normal at the points (4a, 4a) and at the upper end of the latus rectum of the parabolay2 = 4ax intersect on the same parabola.
5. Find the values of for which the line y = x cos + 4 cos3 – 14 cos – 1 is a normal to the parabolay2 = 16x.
6. RULE FOR TRANSFORMINGAN EQUATION FOR THE VARIOUS FORMS OFTHEPARABOLAIn all the previous articles on the parabola, all the related propositions have been proved and derivedfor the particular parabola y2 = 4ax.However, all the results with slight transformations are valid for any parabola. In this Art.If any equation derived for the parabola y2 = 4ax, (a > 0) is given by
E(x, y, a) = 0 ......(1)then the same equationfor the parabola y2 = – 4ax will be
E(x, y, – a) = 0 .......(2)For the parabola x2 = + 4ay will be
E(y, x, a) = 0 ........(3)and for the parabola x2 = – 4ay will be
E(y, x, – a) = 0 ........(4)If the coordinates of the vertex b (), then substitute (x – ) and (y – ) for x and y respectively.
Page 11 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Using the above rule for the equation of general tangent to the parabola y2 = – 4ax, equation is
y = mx –m
a, .........(5)
to the parabola (y – )2 = 4a(x – ) is
(y – ) = m(x – ) +m
a, .......(6)
to the parabola x2 = 4ay is
x = my +m
a.......(7)
Similarly, equation of a general normal to the parabola y2 = – 4ax isy = mx + 2am + am3, .......(8)
to the parabola x2 = 4ay isx = my – 2am – am3 .......(9)
and so on.
Illustration 11:Find the equation of a line which touches the parabola 9x2 + 12x + 18y – 14 = 0 and passes throughthe point (0, 1).
Solution :Equation of the given parabola can be written as
9x2 + 12x + 4 + 18y – 18 = 0i.e. (3x + 2)2 = – 18 (y – 1)
i.e.2
3
2x
= – 2(y – 1) .........(1)
Equation of the tangent to the above parabola can be written as
x +3
2 = m(y – 1) –
m2
1
2
1a2a4 ........(2)
If the tangent passes through (0, 1), then we have0.m2 – 4m – 3 = 0
gives m = – 3/4, Hence, equation of the required lines, are
x +3
2 = –
4
3 (y – 1), x +
3
2 = (y – 1)
i.e. 12x + 9y –1 = 0 and y –1 = 0
Illustration 12:P and Q are the points t
1 and t
2 on the parabola y2 = 4ax. If the normals to the parabola at P and Q
meet at R, (a point on the parabola, show that t1t2 = 2.
Solution :Let the normals at P and Q meet at R(at2, 2at).
Page 12 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Hence t = – t1 –
1t
2 and t = – t
2 –
2t
2
Therefore2
21
1 t
2t
t
2t (t
1– t
2) =
21
21
tt
)tt(2 t
1t2 = 2
Illustration 13:Find the focal distance of the point (x, y) on the parabola x2 – 8x + 16y + 16 = 0
Solution :x2 – 8x + 16y + 16
i.e. (x – 4)2 = – 16y + 16i.e. (x – 4)2 = – 16(y – 1) .......(1)The focal distance of any point (x, y) in the parabola y2 = 4ax is
SP = |x + a|Therefore, using the rules of this Art. The focal distance of any point (x, y) on the parabola x2 = 4aywill be
SP = |y + a|on the parabola x2 = – 3ay will be
SP = |y – a|on the parabola (x – )2 = – 4a(y – ) will be
SP = |y – – a|Therefore, the focal distance of any point (x, y) on the given parabola is
SP = |y – 1– a|= |y – 1 – 4| [ 4a = 16 a = 4]= |y – 5|
Illustration 14:Three normals are drawn from the point (7, 14) to the parabola
x2 – 8x – 16y = 0.Find the coordinates of the feet of the normals.
Solution :The equation of the given parabola is
x2 – 8x – 16y = 0Differentiating above equation throughout w.r.t. x, we have
2x – 8 – 16dx
dy = 0
givesdx
dy =
8
4x
If (h, k) be a point on the given parabola, then we have
k =16
h8h2
and slope of the normal at (h, k) =h4
8
(using result (2))
Therefore, equation of the normal at (h, k) is
Page 13 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
y – k =
h4
8 (x – h)
If the normal passes through (7, 14), then we have
14 –16
h8h2 =
h4
8 (7 – h)
i.e. (4 – h) (224 – h2 + 8h) = 128(7 – h)i.e. h3 – 12h2 – 64h = 0i.e. h(h2 – 12h – 64) = 0gives h = 0, – 4, 16Putting the values of h in equation (3) gives
k = 0, 3, 8respectively.Therefore, the coordinates of the feet of the normals are
(0, 0), (– 4, 3) and (16, 8).
7. EQUATION OFTHE CHORD WHOSE MID-POINT IS GIVENLet P(x
1, y
1) be a given point and
y2 = 4ax ......(1)be a given parabolaEquation of any line passing through P(x
1, y
1) can be written as
y – y1 = m(x – x
1) ......(2)
This line meets the given parabola in point A, B whose abscissa are given by the equation{y
1 + m(x – x
1)}2 = 4ax
i.e. {mx + y1– mx
1}2 = 4ax
i.e. m2x2 + {2m(y1 – mx
1) – 4a}(y
1 – mx
1)2 = 0
If x2, x
3 be the roots of the equation, then according to the given condition, we havex
2 + x
3 + 2x
1[P(x
1, y
1) is the mid-point of AB]
ie. 211
m
)mxy(m2a4 = 2x
1[from equation (3)]
gives m =1y
a2
Putting the above value of m in equation (2) gives the equation of the required chord as(y – y
1) y
1 = 2a(x – x
1)
which on rearranging reduces toyy
1 – 2a(x + x
1) = y
12 – 4ax
1.......(4)
1ST
8. CHORD OF CONTACTLet P(x
1, y
1)be a given point and
y2 = 4ax .......(1)be a given parabola.
Let us assume the coordinate of the point of contact A and B (see fig.) as (x2, y
2) and (x
3, y
3).
Equations of the tangents are A and B (by Art. 6.7) are
Page 14 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
yy2 – 2a(x + x
2) = 0 .......(2)
and yy3 – 2a(x + x
3) = 0 .......(3)
Since these tangents pass through P(x1, y
1), hence its coordinates must satisfy the equations (2)
and (3)i.e. y
1y
2 – 2a(x
1 + x
2) = 0 .......(4)
and y1y
3 – 2a (x
1 + x
3) = 0 .......(5)
Observing equations (4) and (5) it can be seen that the equation of AB isyy
1 – 2a(x + x
1) = 0 ........(6)
Since equations (4) and (5) are true, it follows that A(x2, y
2) and B(x
3, y
3) will satisfy equation
(6), which proves therefore, that equation (6) represents a straight line passing through A, B andhence the chord of P is
yy1 – 2a(x + x
1) = 0
0T
Illustration 15:Find the locus of the mid-point of the chords of the parabola y2 = 4ax such that tangents at theextremities of the chords are perpendicular.
Solution :Let (x
1 , y
1) be the mid-point of the chords y2 = 4ax. Its equation is
yy1 – 2a(x + x
1) = y
12 – 4ax
1
or yy1 – 2ax = y
12 – 2ax
1... (1)
The tangents at the extremities of the chord are perpendicular to each other. Hence the chord is afocal chord. Hence (I) must pass through the focus (a, 0), so that –2a2 = y
12 – 2ax
1
Hence locus of (x1, y
1) is y2 = 2a(x – a).
Drill Exercise - 7
1. Find the equation of the chord of contact of tangents drawn from a point (3, 4) to the parabola y2 = 2x.
2. Find the coordinates of the middle point of the chord of the parabola y2 = 8x the equation of which is2x – 3y + 8 = 0.
3. If the straight line y = mx + c touches the parabola x2 = 4y, then find c in terms of m.
4. Three normals are drawn from the points (14, 7) to the curve y2 – 16x - 8y = 0 find the coordinatesof the feet of the normals.
5. Tangents are drawn at the points where the line lx + my + n = 0 is intersected by the parabolay2 = 4ax. Find the point of intersection of tangents.
9. PAIR OFTANGENTS FROM A GIVEN POINT TOA GIVEN PARABOLALet P(x
1, y
1) be a given point and
y2 = 4ax .......(1)be a given parabola.Let M(h,k) be any point on either of the tangents from P on to the given parabola. The equation of thestraight line joining P and M is
Page 15 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
y – y1 = )xh(
)yk(
1
1
(x – x1)i.e. y = x )xh(
)yk(
1
1
+ )xh(
)kxhy(
1
11
.......(2)
Since this line is a tangent to the given parabola, therefore it must be of the form
y = mx +
m
a.......(3)
Comparing equations (2) and (3), we have
m = )xh(
)yk(
1
1
......(4)
andm
a = )xh(
)kxhy(
1
11
.......(5)
Eliminating m by multiplying equations (4) and (5), we havea(h – x
1)2 = (k – y
1)(hy
1 – kx
1)
Putting (x, y) in place of (h, k) gives the equation of the locus of M (i.e. the required pair of tangents)as a(x – x
1)2 = (y – y
1)(xy
1 – yx
1)
which on rearranging reduces to {yy1 – 2a(x + x
1)}2 = (y2 – 4ax) (y
12 – 4ax
1)
12 SST
10. SOME STANDARD PROPERTIES OFTHE PARABOLA
(i) The portion of a tangent to a parabola intercepted between the directrix and the curve subtends aright angle at the focus :
The equation of the tangent to the parabola y2 = 4ax at P(at2, 2at) isty = x + at2 ......(1)
Let (1) meet the directrix x + a = 0 at Q
Then coordinates of Q is
t
)1t(a,a
2
, also focus S is (a, 0)
Slope of SP =aat
0at22
=
1t
t22
= m1 (say)
Q
Zx+a=0
V
Y
S(a,0)X
Directrix
P(at ,2at)2
90º
and Slope of SQ =aa
0t
aat2
=t2
1t 2
= m2 (say)
m1m
2 = – 1
i.e., SP is perpendicular to SQ i.e., PSQ = 90º.
(ii) The tangent at any point P of a parabola bisects the angle between the focal chord through P and theperpendicular from P to the directrix.Let the tangent at P (at2, 2at) to the parabola y2 = 4ax meet the axis of the parabola.i.e. x-axis or y = 0 at T.The equation of tangent to the parabola y2 = 4ax at P(at2, 2at) is
ty = x + at2
T(– at2, 0) ST = SV + VT = a + at2 = a (1 + t2) Z
x+a=0V S(a,0)
X
Directrix
P(at ,2at)2
M
T
Page 16 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Also SP = PM= a + at2
= a (1 + t2) SP = ST i.e., STP = SPTBut STP = MPT (alternate angles) SPT = MPT.
(iii) The foot of the perpendicular from the focus on any tangent to a parabola lies on the tangent at thevertex.Equation of tangent at P (at2, 2at) on the parabola y2 = 4ax is ty = x + at2
or x – ty + at2 = 0 ......(1)Now the equation of line through S(a, 0) and perpendicular to (1) is Equation tx + y = ta or t2x + ty – at2 = 0By adding equation (1) and (2), we get
x(1 + t2) = 0or x = 0 ( 1 + t2 0)Hence the point of intersection of (1) and (2) lies on x = 0 i.e., on y-axis (which is tangent at the vertexof parabola).
(iv) If S be the focus of the parabola and tangent and normal at any point P meet its axis in T and Grespectively, then ST = SG = SP.Let P (at2, 2at) be any point on the parabola y2 = 4ax, then equation of tangent and normal at P(at2, 2at) are
ty = x + at2
and y = – tx + 2at + at3 respectively.Since tangent and normal meet its axis in T and G. Co-odinates of T and G are (– at2,0) and
(2a + at2, 0) respectively. SP = PM = a + at2
SG = VG – VS = 2a + at2 – a Z
x+a=0
V S
Directrix
P(at ,2at)2
M
T(a,0)
G
90º
= a + at2
and ST = VS + VT = a + at2
Hence SP = SG = ST
(v) If S be the focus and SH be perpendicular to the tangent at P, then H lies on the tangent at the vertexand SH2 = OS. SP where O is the vertex of the parabolaLet P(at2, 2at) be any point on the parabola.
y2 = 4ax ......(1)then tangent at P (at2 , 2at) to the parabola (1) is
ty = x + at2
It meets the tangent at the vertex i.e., x = 0 Co-ordinate of H is (0, at)& SP = PM = a + at2
x+a=0 Y
S(a,0)X
Directrix
P(at ,2at)2
M
O
H
OS = a
& SH = 22 )at0()0a( = 222 taa or (SH)2 = a. {a (1 + t2)} = OS. SP.
(vi) The portion of a tangent to a parabola cut off between the directrix and the curve subtendsright angle at the focus.
Page 17 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
(vii) The tangents at the extremities of a focal chord intersect at right angles on the directrix, andhence a circle on any focal chord as diameter touches the directrix. Also a circle on any focalradii of a point P(at2, 2at) as diameter touches the tangent at the vertex and intercepts a chordof length a t1 2 on a normal at the point P..
(viii) Any tangent to a parabola and the perpendicular on it from the focus meet on the tangent atthe vertex.
(ix) If the tangents at P and Q meet in T, then :
TP and TQ subtend equal angles at the focus S.
ST2 = SP . SQ
The triangles SPT and STQ are similar.
(x) Tangents and Normal at the extremities of the latus rectum of a parabola y2 = 4ax constitutea square, their points of intersection being (-a, 0) and (3a, 0).
(xi) Semi latus rectum of the parabola y2 = 4ax, is the harmonic mean between segments of any focal
chord of the parabola is ; 2a =2bcb c
i.e1 1 1b c a
.
(xii) The circle circumscribing the triangle formed by any three tangents to a parabola passesthrough the focus.
(xiii) The orthocenter of any triangle formed by three tangents to a parabola y2 = 4ax lies on thedirectrix and has the coordinates - a, a(t
1 + t
2 + t
3 + t
1t2t3).
(xiv) The area of the triangle formed by three points on a parabola is twice the area of the triangleformed by the tangents at these points.
(xv) If normal drawn to a parabola passes through a point P(h, k) then k = mh - 2am - am3 i.e.
am3 + m(2a - h) + k = 0. Then gives m1 + m
2 + m
3 = 0 ; m
1m
2 + m
2m
3 + m
3m
1 =
2a ha
;
m1m
2m
3 =
ka . Where m
1, m
2 and m
3 are the slopes of the three concurrent normal.
Note that the algebraic sum of the : slopes of three concurrent normal is zero.
ordinates of the three co-normal points on the parabola is zero.
centroid of the formed by three co-normal points lies on the x-axis.
11. REFLECTION PROPERTY OFA PARABOLAThe tangent (PT) and normal (PN) of the parabola y2 = 4ax at P are the internal and external bisectorsof SPM and BP is parallel to the axis of the parabola and BPN = SPN.
Page 18 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Drill Exercise - 8
1. If the normals at point P(at12, 2at
1) on the parabola y2 = 4ax meet on the parabola, then prove that
t1t2 = 2.
2. From a point P of the parabola y2 = 4ax are drawn the perpendicular PN to the x-axis and the normalPG cutting the x-axis at G. Prove that the length NG is the same for all position of P the parabola.
3. Prove that if a chord of the parabola y2 = 4ax subtend a right angle at the vertex, the tangents at itsextremities meet on the line x + 4a = 0.
4. Prove that the circle described on any focal chord of a parabola as diameter touches the directrix.
5. Prove that any three tangents to a parabola whose slopes are in harmonic progression enclose atriangle of constant area.
Page 19 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
ANSWER KEYDRILL EXERCISE - 1
1. (i)
vertex :
2
5,
2
7, axis : y =
2
5, focus :
2
5,
4
17, directrix : x = –
4
11, latusrectum = 3
(ii)
vertex : (1, 2), axis : x = 1, focus :
4
9,1 , directrix : y =
4
7, latusrectum = 1
2. 16x2 + y2 + 8xy + 96x – 554y – 1879 = 0
3. 4x2 + y2 – 4xy+ 8x + 46y – 71 = 0, axis: 2x –y = 3, LR = 54 unit
4. 16x2 + 9y2 + 24xy – 12x + 16y – 4 = 0
5. (3x + 4y – 4)2 = 20 (4x – 3y + 7)
DRILL EXERCISE - 2
1. (9, ± 6) 2. No such points is possible
Page 20 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
DRILL EXERCISE - 3
1. parabola 2. –1 < m < – 5 + 2 5 3. a (–1, 1)
4.
2
3,
6
5,
25. a (–2, –2 + 2 )
DRILL EXERCISE - 4
2.
4
3tan,
21
DRILL EXERCISE - 5
1. y = x + 3, (3, 6) 2. 4b2 = 7a2 + 2a (c + 2b)
3. 3x -2y + 4 = 0 ; x - y + 3 = 0 4. 2x - y + 2 = 0, (1, 4) ; x + 2y + 16 = 0 , (16, -16)
DRILL EXERCISE - 6
2. Tangent : y = 2x + 2, Normal: x + 2y = 9 ; IInd Tangent : x + 2y + 16 = 0
5. =3
n2 ±
9
n Z
DRILL EXERCISE - 7
1. 4y = x + 3 2. (5, 4) 3. c = – m2
4. (0, 0), (8, 16), (3, –4) 5.
am2
,n
SOLVED SUBJECTIVE EXAMPLESExample 1 :
Show that the line x cos + y sin = p touches the parabola y2 = 4ax if p cos + asin2 = 0 and thatthe point of contact is (a tan2 , – 2a tan)
Solution :The given line isx cos + y sin = por y = – x cot + p coses
Page 21 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
m = – cot and c = p cosec since the given line touches the parabola
c =m
a
or cm = aor (p cosec ) (– cot = a
and point of contact is
m
a2,
m
a2
i.e.
cot
a2,
cot
a2
or (atan2 , – 2atan)
Example 2 :
Prove that the line x
+m
y = 1 touches the parabola y2 = 4a (x + b) if m2 ( + b) + a2 = 0.
Solution :The given parabola isy2 = 4a (x + b) .......(1)Vertex of this parabola is (– b, 0)Now shifting (0, 0) at (– b, 0) thenthen x = X + (– b) and y = Y + 0 ........(2)or x + b = X and y = Y ........(3)From (1), Y2 = 4aX
and the line x
+m
y = 1
reduces to bX
+m
Y = 1
or Y = m
bX1
Y =
m
X + m
b
1 .......(4)
The line (4) will touch the parabola (3), if
m
b
1 =
ma
or
2m
b
1 = – a
or m2 ( + b) + a2 = 0
Page 22 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Example 3 :Find the equations of the straight lines touching both x2 + y2 = 2a2 and y2 = 8ax.
Solution :The given curves are
x2 + y2 = 2a2
and y2 = 8axThe parabola (2) is
y2 = 8axor y2 = 4(2a)x Equation of tangent of (2) is
y = mx +m
a2
or m2 x – my + 2a = 0 ......(3)It is also tangent of (1), then the length of perpendicular from centre of (1) i.e. (0, 0) to (3)
must be equal to the radius of (1) i..e., 2a .
222 m)m(
a200
= a 2
or 224
2
a2mm
a4
or m4 + m2 – 2 = 0or (m2 + 2) (m2 – 1) = 0 m2 + 2 0 (gives the imaginary values) m2 – 1 = 0 m = ± 1Hence from (3) the required tangents are
x ± y + 2a = 0.
Example 4 :Tangents are drawn from the point (x
1, y
1) to the parabola y2 = 4ax, show that the length of their chord
of contact is
|a|
1)a4y)(ax4y( 22
1121 .
Solution :Given parabola is
y2 = 4ax ......(1)Let P (x
1, y
1)
Let the tangent from P touch the parabola at Q (at1
2 , 2at1) and R(at
22, 2at
2) then P is the point of
intersection of tangents
x1 = at
1t2 & y
1 = a(t
1 + t
2) or t
1t2 =
a
x1 and t1 + t
2 =
a
y1
Page 23 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Now QR = 221
222
21 )at2at2()atat(
= ]4)tt[()tt(a 221
221
2 = |a| |t1 – t
2| }4)tt{( 2
21
= | a |
a
x4
a
y 12
21
4
a
y2
21
{from equation 2}
= |a||a|
)ax4y( 121
.|a|
)a4y( 221 P (x , y )
11
R
Q
Chord ofContact
(at , 2at )1 12
(at , 2at )22
2
= |a|
1)a4y)(ax4y( 22
1121 .
Example 5 :A ray of light is coming along the line y = b from the positive direction of x-axis and strikes a concavemirror whose intersection with the x-y plane is a parabola y2 = 4ax. Find the equation of the reflectedray and show that it passes through the focus of the parabola. Both a and b are positive.
Solution :
A ray of light along y = b intersects the parabola at
b,
a4
bP
2
Equation to the parabola is y2 = 4ax ... (1)
Differentiating (1)dx
dy = y
a2
Slope of normal at P isa2
b
If the angle between ray of light and the normal is then tan =a2
b
Let PQ be reflected ray
Angle between PQ and the normal at P is also Let the slope of PQ be m
tan =m
a2b
1
a2b
m
a2
b
ma2
b1
ma2
b
a2
bmm
a4
b
a2
b2
2
XNormal
Y
SA
y=bP
b,
a4
b2
Page 24 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
a2
b2
a4
ba4m
2
22
m = – 22 ba4
ab4
Equation to the reflected ray is y – b = –
a4
bx
ba4
ab4 2
22
a4
bax4
ba4
ab4by
2
22
4abx + (5a2 – b2) y – 4a2 b = 0Substitute S(a, 0)
Reflected ray passes through the focus.
Example 6 :Find the equation of axis, vertex, directrix, extremities of latus rectum and length of latus rectum of theparabola x2 – 3y + 2x + 3 = 0.
Solution :x2 –3y + 2x + 3 = 0
x2 + 2x + 1 = 3y – 2 (x + 1)2 = 3
3
2y . Its axis is the line x + 1 = 0. It’s vertex is
3
2,1 . It’s focus is x + 1 = 0, y –
3
2 =
4
3 i.e.,
12
17,1 . It’s directrix is the line y –
3
2= –
4
3, i.e.,
y +12
1 = 0. It’s latus rectum is the line y –
3
2 =
4
3. i.e., y –
12
17 = 0. Extremities of it’s latus rectum
are x + 1 =2
3, y–
3
2, i.e.,
12
17,
2
1. Length of it’s latus rectum is 3 units.
Example 7 :If the line ax + by + c = 0 is a tangent to the parabola y2 = 4ax then find the corresponding point ofcontact. If the line ax + by + c = 0 is a normal to the parabola y2 = 4ax then find the foot of its normal.
Solution :Let the point of contact of tangent be P
1(at
12, 2at
1). Lines ax + by + c = 0 and yt
1 – x – at
12 = 0 must
be identical,
b
t1 =a
1 =
c
at 21
t1 = –
a
b.
Thus P1
b2,
a
b2
. If foot of normal be P2(at
22, 2at
2) then y + t
2x – 2at
2 – at
23 = 0 and
ax + by + c = 0 must be identical,
b
1 =
a
t 2 =c
)atat2( 322
t2 =
b
a
Page 25 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Thus P2
b
a2,
b
a 2
2
3
.
Example 8 :Prove that the area of the triangle formed by three points on a parabola is twice the area of the triangleformed by the tangents at these points.
Solution :Let the three points on the parabola be at (
121 at2,at ),
22at( , 2at
2), (at
32, 2at
3). The area of the formed by these points
1at2at
1at2at
1at2at
|2
1
323
222
121
| = |a2(t1 – t
2) (t
2 – t
3) (t
3 – t
1)|
The intersection of the tangents, at these points, are the points{at
1 t
2, a(t
1 + t
2)}, {at
2t3, a(t
2 + t
3)}, {at
3t1, a(t
3 + t
1)}
The area of the formed by these points =2
1|a2(t
1 – t
2) (t
2 – t
3) (t
3 – t
1)|.
Example 9 :Find the locus of the point of intersection of two normals to a parabola which are at right angles to oneanother.
Solution :The equation of the normal to the parabola y2 = 4ax is
y = mx – 2am – am3
It passes through the point (h, k) ifk = mh – 2am – am3 am3 + m(2a - h) + k = 0 ... (1)
Let the roots of the above equation be m1, m
2 and m
3.
Let the perpendicular normals correspond to the values of m1 and m
2so that m
1 m
2 = –1.
From the equation (1), m1 m
2 m
3 =
a
k
Since m1 m
2 = –1, m
3=
a
k
Since m3 is a root of (1), we have a
a
k
a
k3
(2a - h) + k = 0
k2 + a(2a - h) + a2 = 0 k2 = a(h – 3a)Hence the locus of (h, k) is y2 = a(x – 3a).
Example 10 :
Page 26 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Three normals from a point to the parabola y2 = 4ax meet the axis of the parabola in points whoseabscissa are in A.P. Find the locus of the point.
Solution :The equation of any normal to the parabola is
y = mx – 2am – am3
It passes through the point (h, k) if
am3 + m (2a - h) + k = 0 ... (1)
The normal cuts the axis of the parabola viz., y = 0 at point where x = 2a + am2
Hence the abscissa of the points in which the normals through (h, k) meet the axis of the parabola are
x1 = 2a + am
12 , x
2 = 2a + am
22, x
3 = 2a + am
32
Since x1, x
2, x
3 are in A.P.
(2a + 21am ) + (2a + 2
3am ) = 2 (2a + 22am )
22
22
21 m2mm ... (2)
Also, from (1), m1 + m
2 + m
3 = 0, ... (3)
m2m
3 + m
3m
1 + m
1m
2 =
a
ha2 ... (4)
and m1 m
2 m
3 = –
a
k... (5)
From (3),
(m1 + m
3)2 = 2
2m 23
21 mm + 2m
1 m
3 = 2
2m
2
22
22
2
22 am
k2mm
am
k.2m2
k2am32
Since m2 is a root of (1), am
23 + m
2(2a – h) + k = 0
2k + m2 (2a – h) + k = 0
{m2(h–2a)}3 = 27k3
a
k2 (h – 2a)3 = 27k3
27 ak2 = 2 (h – 2a)3.Hence the locus of (h, k) is 27 ay2 = 2(x – 2a)3 .
Example 11 :P, Q are the points t
1, t
2 on the parabola y2 = 4ax. The normals at P, Q meet on the parabola.
Show that the middle point of PQ lies on the parabola y2 = 2a(x + 2a).Solution :
The points P and Q are (at12, 2at
1) and (at
22, 2at
2)
As the normals at t1& t
2 meet on the parabola, t
1t2 = 2 ... (i)
Also if (x1, y
1) be the midpoint of PQ, then
x1 =
2
1 (at
12 + at
22) and y
1 =
2
1 (2at
1 + 2at
2) ... (ii)
From (iii) we get (t1 + t
2)2 = (y
1/a)2
Page 27 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
(y1/a)2 = t
12 + t
22 + 2t
1t2 = (2x
1/a) + 4, using (i) and (ii)
y12 = 2a (x
1 + 2a)
Hence the locus of (x1, y
1) is y2 = 2a(x + 2a).
Example 12 :If a circle intersects the parabola y2 = 4ax in four points then show that the algebraic sum of theordinates is zero. Also show that the line joining one pair of these four points and the line joining theother pair have slopes equal in magnitude.
Solution :Solving y2 = 4ax and x2 + y2 + 2gx + 2fy + c = 0 we have y
1 + y
2 + y
3 + y
4 = 0 (y
1, y
2, y
3, y
4 are
ordinates of the points of intersection) y1 + y
2= – (y
3+ y
4)
Slope of line joining (x1, y
1) and (x
2, y
2) is m (say)
2121
22
12
12
12
yy
a4
a4y
a4y
yy
xx
yym
Slope of line joining (x3 y
3) and (x
4 , y
4) is m (say)
4323
24
34
34
34
yy
a4
a4y
a4y
yy
xx
yym
Since y1 + y
2 = –(y
3 + y
4),
We have m = – m Slopes are equal in magnitude.
Example 13 :Tangents are drawn to the parabola at three distinct points. Prove that these tangent lines alwaysmake a triangle and that the locus of the orthocentre of the triangle is the directrix of the parabola.
Solution :Let the three points be A(at
12 , 2at
1), B(at
22, 2at
2), C(at
32, 2at
3).
Tangents at these point are t1y = x + at
12 , t
2y = x + at
22 , t
3y = x + at
32 .
Since t1, t
2, t
3 are distinct, no two tangents are parallel or coincident. Hence these tangents will form
a triangle. The vertices of the triangle are [at1 t
2, a(t
1 + t
2)], [at
2t
3, a(t
2 + t
3)] and
[at3t1, a(t
3 + t
1)]. Equation of the two altitudes are
[y – a(t2 + t
3)] = –t
1 [x - at
2t3] ... (1)
and [y – a (t3 + t
1)] = –t
2 [x – at
3 t
1] ... (2)
Subtracting (2) from (1), we get x = –aHence the locus of the orthocentere is x + a = 0 which is the directrix of the parabola.
Example 14 :Tangents are drawn to a parabola from a point T. If P, Q are the points of contact, prove that theperpendicular distances from P, T and Q upon any tangent to the parabola are in G.P.
Solution :Let the equation of the parabola be y2 = 4ax. Let P(at
12, 2at
1) and Q(at
22, 2at
2) be the two points of
contact. Then the coordinates of T will be (at1 t
2, a(t
1 + t
2)). The equation of the tangent to the
parabola at any point (at2, 2at) is ty = x + at2. Let p1, p
2, p
3 be the length of the perpendicular on this
tangent from the points P, T and Q respectively. Then
|t1
)tt(a||
t1
att2atat|p
2
21
2
122
11
Page 28 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
|t1
)tt(a||
t1
att2atat|p
2
22
2
222
23
so that
|)t1(
)tt()tt(a|pp
2
22
21
2
31
=
2
2
212
21
t1
)tt(atattat
= p
22
p1, p
2, p
3 are in G.P.
Example 15 :Three normals are drawn from the point (c, 0) to the curve y2 = x. Show that c must be greater than1/2. One normal is always the x-axis. Find c for which the other two normals are perpendicular toeach other.
Solution :Equation to normal to the parabola
y2 = x is y = mx –2
m –
4
m3
is passing through (c, 0)
0 = cm –4
m
2
1cm
4
m
2
m 33
m = 0 or4
m
2
1c
2
02
1c
0
4
mcesin
2
c 2
1. If c =
2
1, then m = 0
Then only one normal will be there i.e., x-axis.
Since three normals are there, c >2
1
Then
2
1c2
2
1c2 = –1
c –4
1
2
1 c =
4
3
Page 29 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
SOLVED OBJECTIVE EXAMPLESExample 1 :
Equation of parabola having it’s focus at S(2, 0) and one extremity of it’s latus rectum as (2, 3) is(a) y2 = 4(3 – x) (b) y2 = 4(1 – x)(c) y2 = 8(3 – x) (d) y2 = 8(1 – x)
Solution :Clearly the other extremity of latus rectum is (2, – 2). It’s axis is x-axis. Corresponding value of
a =2
02 = 1. Hence it’s vertex is (1, 0) or (3, 0). Thus it’s equation is y2 = 4(x – 1)
or y2 = – 4(x – 3).
Example 2 :Equation of parabola having the extremities of it’s latus rectum as (3, 4) and (4, 3) is
(a)2
3
7x
+
2
3
7y
=
2
2
6yx
(b)2
2
7x
+
2
2
7y
=
2
2
8yx
(c)2
2
7x
+
2
2
7y
=
2
2
4yx
(d) None of these
Solution :
Focus is
2
7,
2
7 and it’s axis is the line y = x. Corresponding value of ‘a’ is
4
1)11( =
4
2.
Let the equation of it’s directrix be y + x + = 0. 2
|4x3| = 2.
4
2 = – 6, – 8
Thus equation of parabola is
2
2
7x
+
2
2
7y
=
2
)6yx( 2
or2
2
7x
+
2
2
7y
=
2
)8yx( 2.
Example 3 :An equilateral triangle is inscribed in the parabola y2 = 4ax, such that one vertex of this trianglecoincides with the vertex of the parabola. Side length of this triangle is -
(a) 4a 3 (b) 6a 3
(c) 2a 3 (d) 8a 3
Solution :If triangle OAB is equilateral then OA = OB = AB. Thus AB will be a double ordinate of the parabola.
Page 30 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Thus AOX = XOB =6
.
Let A = (at1
2, 2at1) then B (at
12 , 2at
1)
mOA
=1t
2 =
3
1
t1 = 32 AB = 4at
1 = 8a 3 units.
Example 4 :If the normal drawn to parabola y2 = 4ax at the point A(at
12, 2at
1) meets the curve again at
B(at22, 2at
2) then
(a) |t2| 2 2 (b) |t
2| 2 2
(c) |t1| 2 2 (d) |t
1| 2 2
Solution :
We have t2 = – t
1 –
1t
2 t
12 + t
1t2 + 2 = 0.
Since ‘t1’ is real, thus t
22 – 8 0 |t
2| 2 2 .
Example 5 :Locus of trisection point of any double ordinate of y2 = 4ax is(a) 3y2 = 4ax (b) y2 = 6ax(c) 9y2 = 4ax (d) None of these
Solution :Let AB be a double ordinate, where A (at2, 2at), B (at2, – at). If P(h, k) be it’s trisection point
then 3h = 2at2 + at2, 3k =4at – 2at t2 =a
h, t =
a2
k3
Thus locus is 2
2
a4
k9 =
a
h, i.e., 9y2 = 4ax.
Example 6 :Tangents drawn to parabola y2 = 4ax at the point A and B intersect at C. If ‘S’ be the focus of theparabola then SA, SC and SB are in(a) A.P. (b) G.P.(c) H.P. (d) none of these
Solution :If A (at
12, 2at
1), B (at
22, 2at
2), C (at
1at
2, a(t
1 + t
2)).
Now SA = a + at12, SB = a + at
22,
SC = 221
2221 )tt(a)atat(
Page 31 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
a 22
21
22
21 tt1tt = a )t1)(t1( 2
221
Clearly SC2 = SA. SB
Example 7 :Double ordinates AB of the parabola y2 = 4ax subtends an angle /2 at the focus of the parabola thentangents drawn to parabola at A and B will intersect at(a) (– 4a, 0) (b) ( –2a, 0)(c) (–3a, 0) (d) None of these
Solution :
Let A (at2 , 2at), B (at2, – 2at). mOA
=t
2, m
OB = –
t
2. Thus
t
2. –
t
2 = – 1 t2 = 4.
Thus, tangents will intersect at (– 4a, 0).
Example 8 :If a normal chord of y2 = 4ax subtends an angle /2 at the vertex of the parabola then it’s slope isequal to
(a) ± 1 (B) ± 2(c) ± 2 (d) none of these
Solution :Let AB be a normal chord, where A (at
12, 2at
1) and B (at
22, 2at
2).
We have t2 = – t
1 –
1t
2 and t
1t2 = 4 t
1t2 = – t
12 – 2 = – 4 t
12 = 2.
Now slope of chord AB =21 tt
2
= – t1 = ± 2
Example 9 :Length of the shortest normal chord of the parabola y2 = 4ax is
(a) a 27 (b) 3a 3
(c) 2a 27 (d) none of these
Solution :
Let AB be a normal chord where A (at12, 2at
1), B (at
22, 2at
2). We have t
2 = – t
1 –
1t
2.
AB2 = [a2(t1 – t
2)2 (t
1 + t
2)2 + 4] = a2
2
111 t
2tt
4
t
421
= 41
321
2
t
)t1(a16
1
2
dt
)AB(d = 16a2
81
31
3211
221
41
t
t4.)t1(]t2.)t1(3[t = 5
1
221
2
t
)t1(32.a (t
12 – 2)
t1 = 2 is indeed the point of minima of AB2.
Thus ABmini
=2
a4 (1 + 2)3/2 = 2a 27 units.
Page 32 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Example 10 :If three distinct and real normals can be drawn to y2 = 8x from the point (a, 0) then(a) a > 2 (b) a > 4(c) a (2, 4) (d) none of these
Solution :Equation of normal in terms of m is y = mx – 4m – 2m3. If it passes through (a, 0) thenam – 4m – 2m3 = 0
m(a – 4 – 2m2) = 0 m = 0, m2 =2
4a .
For three distinct normal, a – 4 > 0 a > 4
Example 11 :An equilateral triangle SAB is inscribed in the parabola y2 = 4ax having it’s focus at ‘S’. If chord ABlies towards the left of S, then side length of this triangle is
(a) 3a(2 – 3 ) (b) 4a(2 – 3 )
(c) 2(2 – 3 ) (d) 8a(2 – 3 )
Solution :
Let A (at12, 2at
1), B (at
22, – 2at
1). We have m
AS = tan
6 aat
at221
1
= –3
1
t12 + 2 3 t
1 = – 1 = 0 t
1 = – 3 ± 2
Clearly t1 = – 3 – 2 is rejected. Thus t
1 = (2 – 3 ). Hence AB = 4at
1 = 4a(2 – 3 ).
Example 12 :Locus of the midpoint of any focal chord of y2 = 4ax is(a) y2 = a(x – 2a) (b) y2 = 2a(x – 2a)(c) y2 = 2a(x – a) (d) none of these
Solution :Let the midpoint be P(h, k). Equation of this chord is T = S
1. i.e., yk – 2a(x + h) = k2 – 4ah. It must
pass through (a, 0) 2a(a + h) = k2 – 4ah. Thus required locus is y2 = 2ax – 2a2.
Example 13 :Slope of the normal chord of y2 = 8x that gets bisected at (8, 2) is(a) 1 (b) – 1(c) 2 (d) – 2
Solution :
Page 33 of 35 www.StudySteps.in
Study
Steps
.in
Parabola
Let AB be the normal chord where A(2t12, 4t
1), B (2t
22, 4t
2). It’s slope =
21 tt
2
We also have
t2 = – t
1 –
1t
2 and 16 = 2(t
12 + t
22), 4 = 4(t
1 + t
2) t
1 + t
2 = 1. Thus slope is 2.
Example 14 :The length of latus rectum of he parabola, whose focus is (a sin2, a cos2) and directrix is the liney = a, is(a) |ra cos2| (b) |4a sin2|(c) |4a cos2| (d) |a cos2|
Solution :Distance of focus from directrix is |a cos 2 – a|. Thus length of latus rectum is |4a sin2|.
Example 15 :In the adjacent figure a parabola is drawn to pass through the vertices B, C and D of the squareABCD. If A(2, 1), C(2, 3) then focus of this parabola is -
(a)
4
11,1 (b)
4
11,2
(c)
4
13,3 (d)
4
13,2
Solution :Clearly AC is parallel to y-axis. It’s midpoint is (2, 2). Thus B (1, 2).Parabola will be in the form of (x – 2)2 = (y – 3).It passes through (3, 2) l = – . Thus parabola is (x – 2)2 = – 1(y – 3).
It focus is x – 2 = 0. y – 3 = –4
1, i.e.,
4
11,2 .
Page 34 of 35 www.StudySteps.in
