Tangents & Normal RDSA

CHAPTER

6 TANGENTS AND NORMALS

6.1 SLOPES OF TANGENT AND NORMAL In chapter 4, we have learnt that the slope of the tangent to a continuous curve y=f(x) at a point P(xvy$ is equal to

i \

^ . Thus, if the tangent to the curve y =/(*) at point . AxvyJ

(xv y^ makes an angle 6 with x-axis, then

tan8 = n\ JiXy^)

If the tangent at P is parallel to x-axis, then

0 = 0 => tan 0 = 0 dx i. .

v J(xv y2)

= 0

If the tangent at P is parallel to y-axis, then

6 = - = > c o t 0 = O=> \fVjXy yj)

= 0.

We know that the normal to a curve at a point P {x\, y\) is a line perpendicular to the tangent at P and passing through P.

Therefore,

Slope of the normal at P = - Slope of the tangent at P

1 'df dx

v y

\dyJP

6.2 EQUATIONS OF TANGENT AND NORMAL We know that the equation of a straight line passing through (xv y^ and having slope m is y - yx - m (x - Xj).

Since tangent and normal to the curve y = f(x) at point

(*i/J/i) pass through P and have slopes

- 1

dl dx u \ M/Vi)

and

respectively. Therefore, the equation of the tangent

dx V Jxlfyl

at (xi,yi) to y=/(x) is

y-y\ = dx (x - x{)

V J(XV yj)

The equation of the normal to y=f(x) at (x\, y\) is

- 1

or,

REMARK 1

y~y i =

y-y i =

dR dx

- (x -

y (^yi)

){xvy J ( X - X { )

V dx v y

= oo, then the tangent at P (xv y^) is parallel to

y-axis and its equation is x = xv

5.16 MATHEMATICS FOR IIT-JEE

REMARK 2 V \dx ip v JP

= 0, then the normal at P (xv y j is parallel to

y-axis and its equation is x = xv

REMARK 3 The equations of tangent and normal to the curve having its paramaetric equations x =f(t)andy=g(t) are given by

y-g w =

and, y-g(t) = d^±{x-f(t)\

(Equation of tangent)

(Equation of normal)

respectively.

ILLUSTRATIVE EXAMPLES

EXAMPLE L The slope of the curve 2y2 = ax2 + b at (1, -1) is - 1. Find a, b SOLUTION The equation of the curve is

...(i) 2 y2 = ax2+ b Differentiating w.r.t. x, we get

41/4^ = J dx dy ax dx ~ 2y

dy

vdxk~ i)

It is given that the slope of the tangent at (1, - 1) is - 1. Therefore,

- § = - ! = > « = 2.

Since the point (1, - 1 ) lies on (i). Therefore

2 (•-l)2 = a(l)2 + b

=> a + b = 2. - Putting a=2 in a + b = 2, we obtain b = 0.

Hence, a = 2 and b = 0.

EXAMPLE2 Find the point on the curve y-x - llx + 5 at which the tangent has the equation y-x - 11. SOLUTION Let the required point be P(xv yx). Since (Xj, y2)

lies on y-x - l lx + 5. Therefore,

Now, yi = x{-llxi +5

y = x - l lx + 5

? = 3X 2 -11 dx

...(i)

dx , , V J(Xy y j = 3 x f - l l

Since the line y-x- 11 is tangent at the point (xl7 y{). There-fore,

Slope of the tangent at (xv y2) = Slope of the line y = x - 11.

iin dx = Slope of the line x - y - 11 = 0

=» 3x2 - 1 1 " - i

=> = 12

= ±2

Now, = 2

y\ = 2 3 -

Xi = — 2

[Using (i)]

[Using (i)] y = ( - 2) - 1 1 ( - 2) + 5 = 19 So, two points are (2, - 9) and (- 2,19). Of these two points

( -2 ,19) does not lie on y = x - 11. Therefore, the required point is (2,-9). EXAMPLE 3 The curve y = ax3 + bx2 + cx + 5 touches the x-axis at P ( - 2,0) and cuts the y-axis at the point Q where its gradient is 3. Find the equation of the curve completely. [IIT1994] SOLUTION We have,

y = ax3 + bx2 + cx+ 5

= 3ax2 + 2bx + c dx

Since the curve y = ax3 + bx2 + cx + 5 touches the x-axis at P ( - 2,0). This means that the curve passes through P ( - 2,0) and x-axis is the tangent at P ( - 2, 0).

0 = - 8a + 4b - 2c + 5

=*8a-4b + 2c = 5 ...(i)

' a l . o P

3a(-2f + 2bx(-2) + c = 0

12a-4b + c = 0 ...(ii)

The curve y = ax+ bx +cx + 5 fleets y-axis at Q.

Putting x = 0 in y = ax3 + bx2 + cx + 5, we get y = 5 Thus, the coordinates of Q are (0,5). It is given that the gradient of the curve at Q is 3.

and, dx v y

J(x v yd dx = 3

JQ

TANGENTS AND NORMALS 6.3

=> 3ax0 + 2bx0 + c = 3 c = 3

Putting c = 3 in (i) and (ii), we get 8a-4b = - 1 and 12a - 4b = - 3

1 3 Solving these two equations, we get a = - - and& = - - .

Substituting the values of fl, fr and c in the equation of the curve, we obtain

y = - ^ x3 - f x2 + 3x + 5 * 2 . 4 as the equation of the curve. EXAMPLE 4 Find the equation of the normal to the curve

y = (1 + x)y + sin"1 (sin2 x) at x = 0. [IIT 1993] SOLUTION We have,

y = (l + xy + sin"1 (sin2*) ...(i) Putting x = 0, we get y = (1 + 0)y + sin"1 (sin20)

=> y = 1. Thus, we have to write the equation of the normal to (i) at

P( 0,1). Differentiating (i) w.r.t. x, we get

^ = eylog(i + *) d { y l 0 g ( l + x))+ . 1 4 j -

rfx dx & v V l - s i n x ^ (sin2 x)

2 sin x cos x cosxl \ l + sin"x

Putting x = 0 and y = 1, we obtain

v y

dx

dx x 0 + l L + 0 yp

= 1

Hence, the equation of the normal at P (0,1) is

y - 1 = - l ( x - 0 ) or, x + y = 1

EXAMPLE 5 Find the equation of the tangent to the curve xm vm

— + *- = lat the point {xvy{). a b SOLUTION We have,

* m yl ^ — + -Z—= 1 am bm

Differentiating both sides with respect to x, we get

vm-1

...(i)

x"* * ym~1 dy m + m ^ ^ -bm dx 0

dx am

/ \ m - 1

dx /

x m -1

a

The equation of the tangent at (xi, y\) is given by

y - j / i = f J \

.dx . , (X-Xj)

or, y - y j -b"

\m- 1 (x -x a )

OT/amyy^-1-amyr^ = -bm x™-1 x + bm x™

or, (bmx™-l)x + {amy™~l)y = bmx^ + amx^

or,

or,

/ _ -1 \ m -1 X + f j / T 1 ! X +

bm

\ y v y

X + y r - 1 !

am X + bm

V J k y

^ y? y = — +— y bm Dividing both sides

by am bm

y = 1

•.• (.Tp y1) lies on (i)

. $ . yT - + — = 1

Thus, the equation of the tangent to (i) at {x\, y\) is

x m -1

a v y

x - + 0 b v y

EXAMPLE 6 Pircd equation of tangent to the curve x 2 / 3 + y273 = a 2 / 3 flf (xv ya). Hence, proi?e f/zaf the length of the portion of tangent intercepted between the axes is constant. SOLUTION We have,

^ / 3 + y2/3 = a2/3

Differentiating both sides w.r.t. x, we get

2 - i / 3 + 2 - i / 3 ^ = 0

...(i)

dx =

X̂ 1/

1/3 -1/3

rfx v y

v-1/3

(*i,yi) yi y

B

•1/3

/ \ l /3 yi I

y' Fig. 6.2


The equation of the tangent at (xy y{) is

y-y\ dy_ dx {x - Xx)

o r / y-yi = -y-y\

f \ ! / 3

y\

V ^ X

x - x

(x - xx)

or, y\'3 vV3

- 1 / 3 , 2 / 3 _ or/ m -1/3 2 / 3 ^ 2 / 3 or, xx1 •yyi =xi +y i

or, xx .-1/3. Wl

-1/3 _ .2/3 •••(ii)

[•.• (xv y{) lies on (i) /. x 2 / 3 + y f / 3 = a1 2/3-j

Suppose the tangent at P (xi, y{) meets the coordinate axes at A and B respectively.

Putting y - 0 in (ii), we get

xx -1/3 _ ,2/3 = a x = a 2/3 1/3

So, the coordinates of A are (a2/3 x / / 3 , 0 ) Now, putting x = 0 in (ii), we get

yy{1/3 = *2/3 y = a2/3yl/3

So, the coordinates of B are (0, a2/3 y j / 3)

AB = V ^ x ^ V + ^ y j ) 2

= V« 4 / 3 xa 2 / 3

= fl, which is a constant.

t-.. ^ 3 + y2/3 = fl2/3]

Hence, the length of the portion of the tangent intercepted between the axes is constant. EXAMPLE 7 Find all the tangents to the curve y = cos (x + y), -2n<x<2n that are parallel to the line x + 2y = 0. [IIT1985] SOLUTION Let the point of contact of one of the tangents be (Xy 1/2). Then, (xv y1) lies on y- cos (x + y)

yx = cos(x1+y1) ...(i)

dx = - sin (x + y). dx

dx v / = - s i n ( x 1 + y 1 ) U +

(*i< ft) 1

djL dx

1 => - - = - s i n ^ + y j )

=> sin(x1+y1) = l

Squaring (i) and (ii) and then adding, we get

cos2 (x1 + y,) + sin2 (x1 + y :) = y t2 +1

=» l = I / i 2 + l

=> y i = 0

Putting yi = 0 in (i) and (ii), we get

cos x2 = 0 and s inxj = 1

=> = [-, -2n<x1<2n]

Hence, the points of contact are (ti/2, 0) and (- 3n/2, 0) The slope of the tangent is ( - 1/2). Therefore, equations of

tangents are

y - o - i 71

X~2 and y - 0 = 3n x + Y

=> 2x + 4y-7t = 0 and 2x + 4y + 37t = 0

EXAMPLE 8 Determine the quadratic curve y = / (x) if it touches the line y-xat the point x = 1 and passes through the point ( - 1, 0). SOLUTION Let the required quadratic curve be

y = ax2 + bx + c ...(i) It passes through ( - 1, 0). Therefore,

0 = a - b + c ...(ii)

Since the line y = x touches (i) at x = 1. Therefore,

(Slope of the tantent at x = 1) = (Slope of the line y = x)

in dx = 1

Jx=l

2a + b = 1 ...(iii) y = ax2 + frx + c = 2ax +

Since the tangents are parallel to the line x + 2y = 0. There-fore,

Slope of the tangent at {xv y{) = (Slope of the line x + 2y = 0

dx I , 2

The equation of the curve is

y = cos (x + y)

Differentiating w.r.t. x, we get

Putting x = 1 in y = x, we get y = 1

Thus, passes through (1,1).

1 = a + b + c

Solving (ii), (iii) and (iv), we get

1 u 1 A 1 a = -,b = — and c = -

Substituting these values in (i), we get

x2 x 1 y = - + - + —as the required quadratic curve.

...(iv)


EXAMPLE 9 Determine the constant c such that the straight line joining the points (0, 3) and (5, -2) is tangent to the curve

c y~x + l' SOLUTION The equation of the line joining (0,3) and (5, - 2) is

or, y - 3 = - x

or, x + y - 3.

c

Suppose it touches the curve y = at (;q, yi).

Now, c

EXAMPLE 10 Prove that all normals to the curve x = a cos t + at sin t,y -a sin t -at cos t

are at a distance a from the origin. [IIT1983] SOLUTION We have,

x = a cos t + at sin t and y = a sin t - at cos t

% = at cos t and ^ = at sin t at at

<k dy dt at sin £ , . => = — = — = tan t dx ax at cos t

dt The equation of the normal at any point t is given by

1 y - (a sin t - at cos t) = - {x - (a cos t + at sin t)}

V =

dl-

x+1

-c dx (Xl +1)2

But, x + y = 3 touches the curve at (x\, y\). Therefore,

Slope of the tangent at (xv y{) = - 1.

tan t cos t y-(asint- at cos t) = - {x - {a cos t + at sin t)}

y sin t - (a sin t - at sin t cos t)

= - x cos t + a cos2 £ + sin t cos t x cos £ + y sin t = a ...(i) Length of the perpendicular from the origin to (i)

10 cos t + 0 sin t - a I

Vcos21 + sin2 t - = a

- 1 ^ - c (x + 1)2

...(i) => (*! + l)2 = C

Since (*i, y{) lies on the line and also on the curve. There-fore,

* i+y i = 3

and, J/i = +1

...(ii)

...(hi)

Hence, all normals to the given curve are at a distance a from the orgin.

EXAMPLE 11 Find the points on the curve 91/2 = x3 where normal to the curve makes equal intercepts with the axes. SOLUTION Let the required point be {xv y{).

The equation of the curve is Since (x\, yi) lies on the curve. Therefore,

From (i) and (iii), we obtain

x1 = ± VcT - 1 and yl = ±

Substituting these values in (ii), we get

±Vc~-l±Vc = 3

=> ± l4c = 4 -

=> ±4c = 2

c = 4

ALITER The line x + y = 3 touches the curve y =

fore,

Now,

9yi2 = r 3 X1

9 ^ = x3

X2

dx 63/ „ 2 f L

(xv yx)

...(i)

X + l There-

dx \ y

Since the normal to the curve at (x\, yi) make equal inter-cepts with the coordinate axes. Therefore,

Slope of the normal = ± 1

1

C 9 x + t = 3 or, x -2x + c-3 = 0 must have equal roots x + l

:. 4 - 4 ( c - 3 ) = 0 = > c = 4.

dx

' i f dx v y

= ± 1

(*i/yi)

= ± 1


= ± 1

x1 = ± 6 y1

x24 = 36 y2

x r = 36 v 9 , v y

[Using (ii)]

V = 4*2

(x1 - 4) = 0

= 0,4

Putting Xi = 0 in (i), we get

9yi2 = 0 => V l = 0

Putting x\ - 4 in (i), we get

9y , 2 = 43

But, the line making equal intercepts with the coordinate axes cannot pass through the origin.

Hence, the required points are (4,8/3) and (4, - 8/3). EXAMPLE 12 Prove that the segment of the normal to the curve x = 2asint + asmt cos2 t, y --a cos31 contained between the coordinate axes is equal to 2a.

SOLUTION We have,

x = 2a sin t + a sin t cos2 t and y = -a cos31

dx dt

o 9 = 2a cos t + a cos t - 2a cos t sin i

= 2a cos t (1 - sin2 t) + a cos3 t

= 3a cos3 t and,

= 3a cos2 t sin t at

dy _ dy/dt dx dx/dt

3a cos t sin t 3a cos3 t

tan t

The equation of the normal at t is given by

y + a cos3 t = - , ] , (x -2asint-a sin t cos2 t) J dy/dx 3 2 y + a cos t = - cot t (x - 2a sin t - a sin t cos t)

3 , cos t , . , . , 2 fN v + a cos £ = - --—- (x -2asmt-asmt cos t) sin f

x cos t + y sin t = 2a sin t cos £

Suppose this normal meets the coordinate axes at A and B respect ively. Then, the coordinates of AandB are A (2a sin t, 0) and B (0,2a cos t) respectively.

AB = W sin2 £ + 4a2 cos2 t = V4^2 (sin2 t + cos2 0 = 2a. Hence, the length of the segment intercepted between the

coordinate axes is constant. EXAMPLE 13 Find the points at which the tangents to the curves y = x3 - x - 1 and y = 3x2-4x + l are parallel. Also, find the equa-tions of tangents. SOLUTION Let y = mx + c be tangent to y = 3x2 - 4x + 1 at point (xy yx).

Then,

3x2 - 4x + 1 - (mx + c) = 0 must have equal roots.

=> 3x2 -x(4 + m) + l - c = 0 must have equal roots

=> (4 + m)2 - 12 (1 - c) = 0

=> m2 + 8m + 4 + 12 c = 0

c = (m2 + 8m + 4) 12

Since y = mx + c touches y = 3x2 - 4x + 1 at (x\, y\). There-fore,

dx \

- m (*ryi)

6x1 -4 = m

m + 4 x i = —

Since, (xv y2) lies on y = 3x2 - 4x + 1. Therefore,

yj = 3 x2 - 4 x1 +1

A2

Vi = 3

3/1 =

m + 4 - 4 m + 4 + 1

m2 - 4 12

Thus, y = mx + c or, y = mx- m2 + 8m + 4 12 is tangent to the

curve y = 3x - 4x + 1 at the point m + 4 m - 4 12

It is given that the two curves have parallel tangents. So, slope of the tangent to the curve y = x3 - x -1 is also m. Let the point of contact of the tangent of slope m be (x2, yi).

Then,

*L dx . . v J(x2, y2)

= m


3X2 - 1 = w

x0 = ±v m + 1 and m > - 1 " ' 3

Now,

(̂ 2/ 3/2) l ie s on y = x3 - x - 1.

3/2 = x2 ~ x2 ~ 1

, m + 1 y2 = ±

m + 1 - 1

m + 1 y2 = - i ± > 3

m + 1 _ ^

y2 = - i±V m + 1

y2 =

3

m - 2

m-2 3

V3 (m +1)

Thus, the points of contact of tangent of slope m to the curve

y-x -x-1 are 3 ' 9 m > -1. The equation of the tangent at this point is

, where

y-y 2 = ' i f dx (x - x2)

- m-2 /r- — f -Jm +1 ^ y +1 + — V 3 ( m + 1) = m x+ V-^—

y = mx ±2V 1 +m 3

EXAMPLE 14 Tangents are drawn from the origin to the curve y = sin x. Proi?e ffta £ £ftdr pom£s 0/ contact lie on the curve

^ W - y 2

SOLUTION Let (ft, A:) be a point of contact of tangents from the origin to the curve y = sin x.

Now, y = sinx dy_ , = cosx dx

= cos h

The equation of the tangent at (h, k) is

y-k = fk dx

(x-h) J(h,k)

or, y-k = cos ft (x - ft)

It passes through (0,0). Therefore,

-k = -hcosh

=> k — h cos h

Also, (ft, A:) lies on the curve. Therefore, k = sin ft

...(i)

...(ii)

From (i) and (ii), we get

fc2-

v y

9 9 = sin ft + cos ft

/z2

=>i t 2 f t 2 +^ = ft2

=> ^ 2 f t 2 = ft2-^

Hence, (ft, /c) lies on the curve x2y2 = x2 - y2. EXAMPLE 15 Find the equations of the tangents drawn to the curve y2 - 2x3 - 4y + 8 = 0/rom £fte pomf (1,2). SOLUTION Suppose the tangent drawn from (1,2) to the curve y2 - 2x3 - 4y + 8 = 0 touches the curve at (ft, fc).

We have,

y2 - 2x3 - 4y + 8 = 0


J dx dx

dy = 3x2

dx ~ y - 2

dx 3ft

k-2 Xh,k)

The equation of the tangent at (ft, /c) is

y- /c = 3ft (x-f t) k-2 It passes through (1,2). Therefore,

2-k = 3ft Jt — 2

(1-f t)

- (k-2)2 = 3ft2 ( l - /z)

=> 3f t 3 -3 f t 2 - /c2 + 4/c-4 = 0 Also, (ft, /c) lies on (i). Therefore,

/c2 - 2ft3 -4k+ 8 = 0 Adding (iii) and (iv), we get

ft3 - 3ft2 + 4 = 0

=> (ft - 2)2 (ft + 1) = 0

=> ft = - 1 , 2 . Putting ft = 2 in (iii), we get

24 - 12 - /c2 + 4/c - 4 = 0

=> A2 - 4/c - 8 = 0

=> k = 2±2V3"

ft = - 1 gives imaginary values of k.

~(i)

...(ii)

...(iii)

...(iv)


Thus, the points are (2,2 ± 2^3). Putting the values of h and k in (ii), we obtain the following

equations of the tangent

y - ( 2 + 2V3) = 2<3(x-2) and y-(2-<3) = -2<3 (x-2).

EXAMPLE 16 Fznd ffe equation of the tangent to x3 = ay1 at the point A (at2, at3). Find also the point where this tangent meets the curve again. SOLUTION The equation of the curve is

x3 = ay2

Differentrating w.r.t. x, we get ^ -

dy _ 3x2

dx lay

...(i)

' i f dx . , V Xat, at)

3 H 2a2 t3

3t 2

The equation of the tangent at (at2, at3) is

y-af = dx U2, a?)

y-at3 = y(*-«f2)

(x - at )

3tx-2y-at = 0 ...(ii)

Let B (ati , ati) be the point where this tangent again meets the curve.

31 xat22- 2atx

3 -at3 = 0

=> 2t13 - 3 t21 +13 = 0

[v Point B lies on (ii)]

(tl-tf(2tl + t) = 0

^ = t or, ^ = -1 /2

Since ^ = £ gives the point A itself. Therefore, ti=-t/2.

Hence, the required point is B

EXAMPLE 17 7/tfze tangent at (a, b) to the curve x3-\-y3 = c3 meet

of zaf 4 7 8

e c a, b,

the curve again in (alf b{), prove that —

SOLUTION The equation of the curve is x3 + y3 = c3. Differntiating w.r.t. x, we get

3x2 + 3y2& = 0 u dx

fy _ X dx

dx \

.2

i 2

(a,b)

If the tangent at (a, b) cuts the curve again at (ay b\), then

a3 + b3 = c3

ai3 + bi3 = c3 -(i)

...(ii) and,

Slope of the tangent = Slope of the line joining (a, b) and («i/ h)

a2 h-b b2 ai~a

From (i) and (ii), we get

...(iii)

...(iv)

a3 + b3 = a3 + b13

=> a3 -a3 = - (b3 - b3)

a1-a b2 + bb1 + b2

a2 + aa1 + a2

b,-b _a2 + aal + a2

^ a \ ~ a ~ b^ + b b ^ b 2

From (iii) and (iv), we get

a2 + aax + a2 a

2

b2 + b\ + b2 ~ b2

=> a? b2 + aa1 b2 + a2b2 = a2 b? + a2 bbx + a2b2

=> a2b2 - a2 b2 = a2 bb-y - b2 aa^

(a\ b + abi) (a\ b - ab{) - ab (iab\ - ba{)

=> b + ab-y = -ab a \ bi a b

EXAMPLE 18 Prove that the portion of the tangent intercepted between the axes at any point on the curve xmyn = am + n is divided at its point of contact in a constant ratio. SOLUTION Let P (xv y^) be an arbitrary point on the curve

xmyn = am + n. Now,

xmyn = am + n

m log x + n log y = (m + n) log a

m n dy - + = 0 x y dx

d}L = _mt dx nx

[Differentiating w.r.t. x]

dx n Xi x ~X*vyx) " 1

The equation of the tangent at (xy y\) is

y~y\ = nx-i (x - x{)


ft(y-yi) _ _ m(x-x1)

V\ ' xi mx ny — + = m + n xl V\ Suppose it meets the coordinate axes at A and B respective-

ly. Then, the coordinates of A and B are given by

(m + n) xx

m ,0 ,B 0, (m + n) y1

PA -Vf^-x, + ( y i - o )2

_ Vn2 x2 + m2 y2

m and,

PB X 2 J (m + n) y i ) | 2

( o - ^ r + j y i - — - — |

_ ^n2 x i 2 +

n

=> PA : PB = n:m.

Hence, P divides AB in a constant ratio.

EXAMPLE 19 Tangent at point Px (other than (0,0)) on the curve y = x3 meets the curve again at P2. The tangent at P2 meets the curve at P3 and so on. Show that the abscissae ojPv P2, Py , Pnform a GP. Also, find the ratio

area A P1 P2 P3

area A P2 P3 P4 [IIT1993]

SOLUTION Let P1 (t, t ) be any point on the curve y = x .

Now, y = x

dx dx = 3£z

At,

The equation of the tangent at (t, t ) is

y - f3 = 312 (X - t)

=> 3t2x-y = 2t3 ...(i)

This meets the curve y = x3 again at P2 whose x-coordinate is given by

3£2 x - x3 = It3 [Putting y = x3 in (i)]

=> x3 -3t2 x + It3 = 0

=> (x-1)2 (x + 20 = 0 => x = - It [•'.• x = t corresponds to point P2]

Putting x = - 2t in y = x3, we obtain

y = (-21)3 = - S t 3

So, the coordinates of P2 are Now,

dx = 3 (- 21)2 = 1212

The equation of the tangnet at P2 (- 21, - St3) is

y + St3 = 12 f2 (x + 2t) The x-coordinates of the point where this tangent meets

with the curve y = x3 is given by

x3 + St3 = 12£2 (x + 2t)

=> x3 - 12£2 x -16£3 = 0

=> (x + 2tf(x-4t) = 0

=> x = 4£ [•.• x = - 2f corresponds to point P2]

Putting x = 4f in y = x3, we get

y = 64 f3

Thus, the coordinates of P3 are Continuing like this, we obtain the coordinates of P4 as

(- St, - 5\2t3). The abscissae of Pi, P2, P3, P4 are

f , - 2 f , 4f , - 8 f

Clearly, they form a GP with common ratio - 2. Now,

Aa = Area of AP1P2P3 = -

and,

A2 = Area of A P 2 P 3 P 4 = |

£ f3 1 -2f -8£3 1 At (At3 1

- 2f - 8 r 1 4f 64f3 1

- 8f - 512£3 1

= - x - 2 x - 8

= 16 Aa.

£ f -2t - St3

41 (At3

= _1_ ~ 16

Area of A P1P2P3 l OX' Area of A P2 P3 P4 16

EXAMPLE 20 Fznd f/ze condition that the line xm \F

x cos a + y sin a = p may touch the curve — + = I.

SOLUTION We have,


xm v" , am bm ...(i)


= 0 mxm~1 mym~l dy dx

dy = -bm

dx nm

/ \m- 1

Kb Suppose the line x cos a + y sin a = p touches (i) at (xlf t/{).

m m

a"1 ...(ii)

and,

'(k dx , V AxvyJ

-bn / \ra -1 f i

A The equation of the tangent to (i) at (x\, y\) is

( y - y i)

y - y i

m-1 yi

Jb (x-x 2 )

1 (x - x2)

y y ^ " 1 yim -XX m-1

x r m - 1 .,. - 1 yy i v m „ m X1 | yi

/ \ X a

v y

(

a v y

h b v y

a

\m -1 = 1 ...(iii) [Using (i)]

If the line x cos a + y sin a = p touches (i) at (x\, y{), then this line and (iii) must coincide.

/ \m-1 11 a

\ y

cos a

h b v y

\W -1

sin a 1 V

\tn-l

a v y

a cos a

0 cos a

P \1/m - 1

and Vi \m -1

fr sin a

and

& y V

yi ft • > fr s m a

I P ;

l /m- l

Subsituting the values of ^ and ~ in (ii), we get

a cos a m / ( m -1)

b sin a \m/(m -1)

= 1

+ (b sin a)m / ( m " « = pm/^m ~ l \ which is the (a cos a)m/{m~l)

required condition. EXAMPLE 21 Find the condition that the line x cos a + y sin a = p may touch the curve xm yn = am + n. SOLUTION Suppose the line x cos a + y sin a = p touches the curve xmyn = am + n at (xv y1}. Therefore,

and

Now,

xx cos a +1/1 sin a = p

x™y\ = am + n

xmyn = am + n

...(i)

...(ii)

m log x + n log y = (m + n) log a

m n dy - + - j = 0 x y dx

dy_ = _rny_ dx n x

dx mVi n x-,'

The equation of the tangent to xm x/1 = am + n at (xlf yi) is

mV i / x y ~y i = (x - x . ) u yL n x- 1

or, (my{)x + (nx{)y = (m + n)x1y1

or, / \ m

l v y

x + f \ n_

v y - m + n.

If the line x cos a + y sin a = p touches xmyn = am + t

/ yi)/ then this should be same as x cos a + y sin a = p. at

*1 A yi m + n

cos a sin a

m n x-, cos a

Xi =

y1 sin a

mp

m + n P

yy\ n p

1 (m + n) cos a ^ (m + n) sin a

Substituting the values of m and n in (ii), we get

mp [ I np _ ^ (m + n) cos a j ] (m + n) sin a

=> rnmnnpm + n = (rn + n)m + nam + n( cos a)m (sin a)n.

This is the required condition.

EXAMPLE 22 Fznd condition that the line Ax + By-1 may be a

normal to the curve an ~1 y = xn.


SOLUTION Let the line Ax + By = \ be normal to the curve an "1 y = Xn at point (x1, y{). Then, (x1, y{) lies on the curve as well as on the line and the slope of the normal at (xv is same as the slope of the line. Now,

a y = rfy = n x" dx

n x"

n / dx

^L = X

m

-ft) xi

.72-1 nxn~l .y ... an~1y = xn

1 xn

/. fl""1 = — y

dx

The equation of the normal at (xv is

y-y i m (x - xa)

2 2 => xx1 + nyyl - nyl + xl

But the normal at (x\, y\) is given to be Ax-+ By = 1. There-fore, comparing these two, we obtain

™/i ny\ + x \ A B

xx = AX, y2

1

BX n

= a , say

and ny 2 + xt2 = X

n B2X2 +A2X2 = X

(B2 + A2 n)X = n

X =

x, =

B2 + A2n

An A - and Vi = 1 B2 + A2n 1 B2 + A2n

Since (x\, y\) lies on the curve an y = xn. Therefore,

an~l B B2 + A2 n

„n-l um2_

An \n

v B2 + A2 n

a" A B (B2 + A n)n = ( . A n f , wh ich is the required condition.

EXAMPLE 23 Prove that the curves y =f (x),(f (x)> 0) and y —f(x) sin x, where f(x) is differentiable function, have common tangents at common points.

SOLUTION The equations of t w o curves are

y = f(x) ...(i)

and, y = f ( x ) s in x ...(ii)

Solving these t w o equations, w e get

/ ( x ) = / ( x ) . s i n x

=> s i n x = 1 [... / ( x ) > 0 ]

x = 2nn + - ' n e Z

x = (4n + l ) - ' n e Z.

Differentiating (i) and (ii) w.r.t. x, we get

^ = f'(x) and ^ =/'(*) sin x + /(x) cos x

mi = Slope of the tangent to (i) at x = (4n + 1) -

7l]

and,

m2 = Slope of the tangent to (ii) at x = (4n + 1) -

= / ' (4n + l ) ~ sin (4n + l ) ^ + / (4n + l ) ~ cos (4rc + l ) ~

= / ' (4n + l ) -

Clearly, m\ = m2. Hence, the two curves have common tan-gents at common points. EXAMPLE 24 Find the equation of a straight line which is tangent to one point and normal to the other point on the curve x = 4t2 + 3,y = 8t3-l.

SOLUTION The parametric equations of the given curve are

x = At2 + 3 and y — St3 — 1

/. ^ = St a n d f = 2 4 f 2

dt dt

NT da dy/dt 24 N™'dx = dxW = -W = 3 t

The equation of the tangent to the given curve at any point t is given by

y - (St3 - 1) = 3t (x - 4£2 - 3) ...(ii)

If this tangent meets the curve again at t\, then

(4£i2 + 3, St3 - 1 ) must satisfy it.

St3 - 1 - (St3 - 1 ) = 31 (4 tx2 + 3-4t2-3)

8 ( t - 1 3 ) = 121 (fa2 -12)

=>4 (tl- t) (2 (tx2 + t2 + ttx) - 31 (tt +1)} = 0


4 ( ^ - 0 (2fj - ftj t2}

4 (fj - ty (2fj + 0 = 0

f j = f or f j = - - .

sec2 9

sec 0 =

f 2 9f2 +1 6 f

v

9t2 +1 6f

Since f j = t gives the point where tangent is drawn. There- •'• t a n 9 + sec 0 =

fore,

Now, (Slope of the normal at fj) = 4m. dx V A'j)

- 1 3 U

[Using (i)]

Since line given in (ii) is normal at f^. Therefore, slope of normal at t\ = slope of (ii)

91 U 9t [•.•«! = ~t/2]

9 ~ 3

Putting the values of t in (i), we obtain the following equations of the required straight lines.

pr 89<2 . , nr 89^2 V2x-y = + \ and V2x+y = - 27 1

EXAMPLE 25 Show that the curve x = 1 - 312, y = t-3t3 is sym-metrical about x-axis and has no real points for x > 1. If the tangent at the point t is inclined at an angle 0 with x-axis, prove that tan 0 + sec 0 = 3f. If the tangent at P ( - 2, 2) meets the curve again at Q, prove that the tangents at P and Q are at right angles. SOLUTION We observe that the expression for x does not change when t is replaced by -1. So, the curve is symmetrical about x-axis. Also, for x > 1, the equation x = 1 - 3f2 gives im-aginary values of f. So, the curve has no real points for x > 1.

Now, x = 1 - 312 and y = t - 3t3

1 -9f ^ = -6t and $ = 1 -<"2 dt dt

dy_ = dy/d dx dx

dt

1 - 9 r -61 ...(i)

tan 0 = 9t2 - 1 61

1 + tan2 0 = 1 +

v The tangent makes an angle 0 with x-axis

9t2 -1 61

9t2 - 1 9f2 + 1 61 + 61

When x = - 2 and y = 2, we have 1 - 3t2 = - 2 and t - 3t3 = 2

=> t = -1 Putting t = - 1 in (i), we have

1 - 9 = _ 4 6 ~ 3

= 3t.

dx v

So, the equation of the tangent at P ( - 2,2),is 4 y - 2 = - - (x + 2) or 4x + 3y + 2 = 0

Suppose this tangent meets the curve at f j i.e. at

Q (1-3^,^-3^)

4(1 - 3fa2) + 3 (fj - 3<!3) + 2 = 0

- 9 f j3 - 12t12 + 3t1 +6 = 0

3 f j 3 + 4fj2 -t1- 2 = 0

(f1 + l ) 2 (3f ! -2) = 0

=> h = " M i 2

- 3'

Clearly, ^ = - 1 corresponds to point P. So, t} = - .

Now,

Wj = Slope of the tangent to the given curve at P

V2 = 9 x ( - l)z - 1 _ - 4

6 x - 1 ~ 3 [Putting f = - 1]

and,

m2 = Slope of the tangent to the given curve at Q

9 x (2/3) - 1 6 x 2 / 3

4 - 1 _ 3 4 ~ 4'

[Putting t = t1=:,

Clearly, mi m2 = - 1 Hence, the tangents at P and Q are at right angles.

EXAMPLE 26 Show that the curve y = x2sin3x touches the parabolas y = x2 and y = - x2 alternately at infinite number of points. SOLUTION We know that two curves touch each other at a point of intersection iff their slopes at the point are equal.


2 First we find the points of intersection of y = x sin 3x and

y = x2. At the points of intersection, we have

x2 sin 3x = x2

=> sin 3x = 1

=> 3x = 2nn + ^

2nn n X = ^ + 6 ' n £ Z

Now,

y = x2 sin 3x

= 2x sin 3x + 3X2 cos 3x ax

rdj? dx i 2nn n

= 2 2 nn n* ~3~ + 6

A nn 7i ...(i)

and,

y = ^

^ = 2x dx r i \ m. dx

V 2rm n atx = — + -

= 2 2nn n T + 6 V

Ann n T + 3 •••(ii)

From (i) and (ii), we find that the curves y = X2 sin 3x and y = x2 have the same slope at their points of intersection. So, they touch each other.

Let us now find the points of intersection of y = x2 sin 3x and y = -x2. At the points of intersection, we have

x sin in3x = -x2

sin 3x = -1

3x = 2nn ~2'ne Z

2nn n „ X = ~ V ~ 6 ' n e Z

Now,

y = x sin3x

dx v atx 2nn _ n 3 ~ 6

= - 2 r2rm n"

and,

y ~ ~x

'dy) dx | , 2nn n fltX = ~ ?

= - 2 2 nn n^

3 6

Ann n

-Ann n 3 + 3

...(iii)

...(iv)

From (iii), (iv), we observe that the curves y = X2, sin 3x and y = -x2 have the same slopes at their points of intersection. Hence, they touch each other.

2nn n It is obvious that the sets of points given by x = —— + — and

2nn n x = —r— - —, n e Z occur alternately.

3 6 Hence, y = x2 sin 3x touches y = x2 and y = - x2 alternately

at infinite number of points. X

EXAMPLE 27 For the function F (x) = J 2 \ t\ dt,find the tangent 0

lines which are parallel to the bisector of the angle in the first quadrant. SOLUTION We have,

X

F (x) = J 2 111 dt

X

J 2 tdt , if x > 0 o X

-J 2tdt,ifx<0

[IIT1995]

F(x)

F(x) =

F'(x) =

Thus,

jx2 , if x > 0 [ - z 2 , if x < 0

(2x , x >0 j - 2 x , if x < 0

CASE I When x > 0

In this case, w e have

J2 y = x dx It is given that the tangent lines are parallel to the bisector

of the angle in the first quadrant.

• ^ = 1 =» 2x = 1 =* X = i dx 2

1 2 1 Putting x = - i n y = xz, we get y =

So, the equation of the tangent at (1 /2 ,1 /4) is

x-: y = x--* A

CASE II When x < 0.

In this case, we have

Since the tangent lines are parallel to the bisector of the angle in the first quadrant.

. <k = dx

l = > - 2 x = l = > x =

1 2 1 Puttingx = - — iny = - x , w e g e t y = - - .


So, the equation of the tangent at ( - 1 /2, - 1 / 4) is a2 = —2a v a2 * a]

1 = or, x - y + - = 0. 3 4

EXAMPLE 28 Tangent at the point P1 = (a, a + 1), (a * 0) on the

curve y = x3 + 1 meets the curve again at P2• The tangent at P2 meets In 2 n

the curve again at P3 and so on. Find E x (P() and £ y (P;), where i=i <=i

x (P;) and y (P;) are the abscissa and ordinate of Pi respectively. • SOLUTION We have,

y = x3 + l

dx

V +1) 3a2.

...(i)

The equation of the tangent at Pi {a, a3 + 1) is

y-(a3 +1) = 3a2 (x -a ) This meets the curve at P^ So, let the coordinates of P2 be

(«2,«2 + 1). As P2 lies on (i), therefore

a23 + 1 - (a3 + 1) = 3a2 (a2 - a)

2

=> a2 -a = 3a (a2 - a)

(a2-a)2 (a2 + 2a) = 0

So, the coordinates of P2 are ( - 2a, - 8a3 + 1) Comparing this with the coordinates of P\, we find that the

coordinates of P2 can be obtained from the coordinates of Pi if we replace a by - 2a.

Similarly, coordinates of P3, P4, . . . are

P3 (4a, (4a)3 + 1), P4 ( - 8a, (- 8a)3 +1) and so on.

Now,

2 n 2 z (P,) = a - 2a + 4a - 8a + ... upto 2n terms

! = 1

= a l - ( - 2 ) 1 - 2

2n

= 1 ( 1 - 4 " )

and, 2n S y (P;) = (a3 +1) + ( - 8a3 +1 + (64a3 +1) + ... 2n terms

i = l

= a* (I - 8 + 64+ ...) + 2n

= y ( l - 6 4 " ) + 2 n.

EXERCISE 6.1

1. Show that the line — + ^ = 1 touches the curve y = he x/" at a b

the point where the curve crosses the y-axis. 2. Find the tangents to the curve y = (x3-1) (x-2) at the points where the curve cuts the x-axis. 3. Find the equations of the tangent and normal to the curve x3 + y3 = 6xy at the point (3,3). 4. Find the equations of the tangent and normal to the curve 2y = 3 - x2 at the point (1,1). 5. Find the equation of tangent and normal to the following curves at the points where ^-coordinates are a :

(i) y 2 (2a -x ) = x3 (ii) y2 (a + x) = x2 (3a - x) 6. Find the equations of the tangent and normal to the curve y (x - 2) (x - 3) - x + 7 = 0 at the point where it cuts the axis of x. 7. Find the equations of the tangent and normal to the curve y = x3- 2x2 + 4 at the point x = 2.

Find the equations of tangents to the curve 3x2 + y2

+ x + 2y = 0 which are perpendicular to the line 4x - 2y = 1. 9. Find the equations of tangent and normal to the curve whose parametric equations are

- u a -x / a : X =

2atz 2 at5

1 + f 2 ' y = 1+ t 2 at the point t = -

10. Show that the sum of the intercepts made on the axes of coordinates by any tangent to the curve Vx7 + Vy~ = V^ is equal tOfl. 11. Find the point on the curve y = x4 - 6x3 + 13x2 - lOx + 5 where the tangent is parallel to the line y = 2x. Show that two of these points have the same tangent. 12. Find the equation of the normal to the curve whose paramet r ic equa t ions are 3 cos ( - cos y = 3 sin 8 - sin 8 at any point 6. Also, prove that the normal at a point 8 = n/4 passes through the origin. 13. If the t angent at any point (1,2) on the curve

2 7 y = ax + bx + — be parallel to the normal at ( - 2,2) on the curve

y-x + 6x + 10, find the values of a and b. 14. Prove that the tangent at any point on the rectangular hyperbola xy = c2, makes a triangle of constant area with the coordinate axes.


15. Prove that the segment of the tangent to xy = c intercepted between the axes of coordinates is bisected at the point of contact. 16. Find the equation of the normal to the curve x3 + y3 = 8xy at the point other than the origin where it meets the curve y1 = 4x. 17. If a, (3 are the intercepts made on the axes by the tangent at any point of the curve x = a cos 8,y = b sin 8, prove that

* + £ = 1 a2 b2

18. Show that the tangent to the curve

= 2. point (a, b) is ̂ + ^

f \n x a v y b v

•- 2 at the

19. Find the abscissa of the point on the curve ay2 - x3, the normal at which cuts off equal intercepts from the axes. 20. Find the equations of tangents to the curve y = x4 which are drawn from the point (2,0). 21. Show that two tangents can be drawn from the point A (2a, 3a) to the parabola y2 = 4ax. Find the equations of these tangents. 22. Find the condition that the line x cos a + y sin a = p may touch the curve

2 2

(ii)

(iii) •a \n/n -1

, yi/n -1 X

a v / yi/n-1

n/n -1

+ , f l - i

= 1

23. Find the abscissa of the point on the curve xy = (x + c) , the normal at which cuts off numerically equal intercepts from the axes of coordinates. 24. For the curve whose paramet r ic equat ions are

a 2 t x = a cos t + — log tan — and y = a sin t, show that the portion

of the tangent intercepted between the point of contact and the x-axis is of constant length. 25. Prove that the portion of the tangent to the curve

y2 + 9 ^ g 1 a + a - ^ - y 2

between the point of contact and the x-axis is of length a. 26. If ax + by = 1 is a normal to the parabola y2 = 4cx, prove that

ca3 + 2 ab3c = b2

27. If x, and y1 be the intercepts on the axes of X an Y cut off

by the tangent to the curve a v /'

^ | = 1, then prove that

, \n/n -1 a

V V Vyi7 = 1.

28. If (x-|, y,) be the parts of the axes intercepted by the tangent s2/3

at any point on the curve / >2/3 'x a

V1

+

b

a = 1, prove that

• 1.

29. The tangent at any point of the curve x = a cos t, ri

y = a sin t meets the axes in P and Q. Prove that the locus of the mid-point of PQ is a circle. 30. If pj and p2 be the lengths of perpendiculars from the origin on the tangent and normal to the curve v-2/3 + y 2 / 3 = a2/3 respectively, prove that 4 p 2 + p2 = a2. 31. Show that the normal to the rectangular hyperbola

xy = c at the point P ctv- meets the curve again at the point

Q ct2,— , if Ud t2 = - 1.

32. Show that the normal to the curve 5x - lOx +x + 2y + 6 = 0 at P (0, - 3) meets the curve again at two points. Find the equations of tangents at these points. 33. If the tangent at the point (at2, at3) on the curve ay2 = xJ

meets the curve again at Q, then prove that the coordinates of

Q are f 2 at at"

34. Find the equation of the straight line which is tangent at one point and normal at another point of the curve x = 3 t2,y = 2t3.

ANSWERS

2. 3x + y = 3,7x-y = 14 3. Tangent: x + y = 6, Normal x - y = 0 4. Tangentx + y = 2, N o r m a l x - y = 0 5. (i) At (a, a) Tangent: 2x - y - a = 0, Normal: x + 2y - 3a = 0

At (a, - a) Tangent :2x + y - a = 0, x-2y-3a = 0 (ii) At (a, a)

Tangent: x - 2y + a = 0, Normal: 2x + y - 3a = 0 At (a, - a)

Tangent :x + 2y + a = 0 , 2 x - y - 3 f l = 0

6. Tangent: x - 20 y - 7 = 0, Normal 20x + y -140 = 0 7. Tangent: 4x - y = 4, Normal: x + 4y = 18 8. 3x + 6y + 13 = 0 , x + 2y = 0 9. Tangent: 13x - 16y = la, Normal: 16x + 13y = 9a 11. (1,3), (2,5) (3/2, 65/6)

13. a = 1, b = -1

19. 4a

16. y-x = 0

20. y = 0, y 3 v y = 4

3 v / X--

13


21. x-y + a = 0, x - 2y + 4a = 0 22. (i) a2 cos2 a + b2 s in 2 a = p2

(ii) (a cos af + (b sin a)" = pn

(iii) (p - a cos a - b sin a)" = (a cos a)n + (b sin a)"

23. x = ± <2 34. y = V2*-2V2 ,y = -<2x + 2<2

HINTS TO SELECTED PROBLEMS

30. Write tangent and normal at any point (a cos 0 , a sin 0) on the curve x 2 / 3 + y2/3 = a2/3

31. Euqate the slope of the normal at P to the slope of line PQ

32. The equation of the normal at P (0, - 3) is x + 2y + 6 = 0. The points of intersection of this normal and the curve are given by 5X5-10X3 = 0 or, x = 0, ± V2. x = 0 gives the point P. Hence, the normal meets the curve again at two points.

— • • •

6.4 ANGLE OF INTERSECTION OF TWO CURVES The angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection.

Let Q and C2 be two curves having equations y =f(x) and y = g(x) respectively. Let PT\ and PT2 be tangents to the curves Cj and C2 respectively at their common point of intersection. Then the angle 0 between PT\ and PT2 is the angle of intersec-tion of Ci and C2. Let v|/i and 1̂ 2 be the angles made by PTj and

Fig. 6.3

PT2 with the positive direction of x-axis in anticlockwise sense. Then

ni\ = tan \|/i = Slope of the tangent to y =f(x) at P

dx _ v / Cj and,

m2 = tan \|/2 = Slope of the tangent to y = g(x) at P

dx yc2

From Fig. 6.3, it follows that

<t> = % -

tan <|> = tan (\|/1 - \|/2)

tan \|/j - tan \f/2

1 + tan tan i|/2 tan d> =

tan 0 = dx

\ y dx \

1 + <k dx

v y

<k dx

The other angle between the tangents is 180° - 0. Generally the smaller of these two angles is taken to be the angle of intersection. ORTHOGONAL CURVES If the angle of intersection of two curves is a right angle, the two curves are said to intersect orthogonally and the curves are called orthogonal curves.

If the curves Q and C2 are orthogonal, then 0 = n/2.

dx m\ dx

C, = - 1

V S<-2 REMARK The two curves Cj and C2 will touch each other at a point

P. if <k dx

K

<k dx


EXAMPLE 1 Find the angle of intersection of the following curves : (i) xy = 6 and x2y = 12 (ii) y2 = 4x and x2 = 4y SOLUTION (i) The equations of the two curves are

xy = 6 ...(i) and x2y = 12 ...(ii)

From (i), we obtain y = ~. Putting this value of y in (ii), we

obtain

= 12 => 6x = 12 x = 2


Putting x = 2 in (i) or (ii), we get y = 3. Thus, the two curves intersect at P (2,3)

Differentiating (i) w.r.t. x, we get

x—L + v = 0 ax y

<k. = dx x m\ <k

dx

Differentiating (ii) w.r.t x, we get

x2 + 2 xy = = =

(2, 3)

dx (2,3) dx ' dx x

Let 0 be the angle of intersection of (i) and (ii), then

-2 - ( 3 / 2 ) + 3 3

= - 3

tan 6 = 1 +m 1 m 2 1 + ( - 3 / 2 ) ( - 3) 11

0 = tan -1 11 V J

(ii) The equations of the two curv.es are

f = 4x and x2 = 4 y ...(ii)

V V From (i), we obtain x = -^p Putting x = ^ in (ii), we get

( A = 4 y => y4 - (Ay = 0 => y(y3 -64) = 0=>y = 0,y = 4 v /

From (i), when y = 0, we get x = 0 and when y = 4, we get x = 4.

Thus the two curves intersect at (0,0) and (4,4). Differentiating (i) w.r.t. x, we get

M - 4 _ i n = 2 2y , - - -v , -ax dx y Differentiating (ii) w.r.t. x, we get

2 x = 4 J x ^ c k ^ x

dx dx 2

Angle of Intersection at (0,0):

...(iii)

...(iv)

From (iii), we have,

m dx v (0,0)

Therefore the tangent to (i) curve at (0,0) is parallel to y-axis. From (iv), we have

(dy) m2 Jx v y

= 0 (0,0)

Therefore the tangent to (ii) curve at (0, 0) is parallel to x-axis.

Hence the angle between the tangents to two curves at (0,0) is a right angle. Consequently the two curves intersect at right angle at (0,0).

Angle of Intersection at (4,4):

From (iii), we have

m, = dx 7(4,4)

2 _ 1 4 " 2'

From (iv), we have,

m2 dy dx v y '(4,4)

Let 0 be the angle of intersection of the two curves. Then 3 4' tan 0 =

- mi 2-(1/2) 1 +m-lm2 1 + 2 x (1/2)

EXAMPLE 2 Find the acute angle between the curves y - 1| and y- • 31 at their points of intersection. SOLUTION Graphs of the two curves y= \x2-l\ and y = | x2 - 31 are shown in Fig. 6.4

y

"C (0,3) u. \ \ \ /T 11 /

" A U \ \ \ \ * \ i>

1 * / * h' / * 0» \ \ j \ / / i,

(0.1) V

\ QX p / x' \ / \ 0 X

(—yfi.O) (--1,0) (1,0) \ (v/3.0)

y = - ( < 2 - i )

v

Fig. 6.4

Clearly, two curves intersect at P and Q. Point P and Q are the intersection point of y = x - 1 and y = -(xr - 3). Solving these two equations simultaneously, we get

x = ± V 2 and y = ± 1. Thus, the coordinates of P and Q are

(V2,l) and ( -^2 ,1) respectively. Angle of intersection at (V271):

The equations of the two curves are

...(i) ...(ii)

[For curve (i)]

and y dy dx

and,

y = x 2 - l

= 2x

= - 2x

(x 2 -3)

dx [For curve (ii)]

Let m\ and m2 be the slopes of the tangents at point P to curve (i) and (ii) respectively. Then,

c j \ <1

\ dy dx v

= 2\2 and m2 = dy dx v

= - 2 \<2

Let 0 be the acute angle of intersection. Then,

tan 0 =

tan (

1 + m1m2

• 2V 2 + 2V2 1 + 2 \ 1 x - 2 V2

4\2 7


, e » , a n ' f f )

Angle of intersection at Q (-

In this case, we have

m1 = Slope of the tangent to (i) curve at Q =

= -2V2

We have, mi = Slope of the tangent at P to (i)

- b 2 y[T= ~f>3/2

a2 b a3/2

and,

\dXjQ

dx

and, m2 = Slope of the tangent at P to (ii) =

Let 6 be the acute angle of intersection at P. Then,

m2 = slope of the tangent to (ii) curve at Q =

= 2V2

So, the acute angle 61 between the tangents is given by

tanGj =

tan 0 = m1 - m2

m1 - m2 - 2 ^ 2 - 2 V 2 1 + fflj m2 1 + {-2<2)Y.{2<2)

4V2 7

1 + ttij m2

(- b2 + a2)

-b3/2 , [7 a 3 ' 2 + h „ &

1 + -a

=> 0j = tan - 1

(1a + b) Jab

b-a Jab

[v b>a]

7 V J

x2 V2

EXAMPLE 3 Find the angle of intersection of 2 = 1 and a b

x2 + y2 = ab. SOLUTION The equations of the two curves are

1 ...(i)

Similarly, we can find the angles of intersection at the remaining points also. EXAMPLE 4 Show that the condition that the curves

„2

and

ax" + by2 = 1

a' x2 + V y2 = 1

...(i)

..(ii)

2 2

a2 b2

and, x2+y2 = ab ...(ii) Solving (i) and (ii) by cross-multiplication, we obtain

x2 _ y2 _ 1 -ab 1 _1 ab 1 b2 a2 a2~b2

a2 b2

b2-a2

1 1 1 1 should intersect orthogonally is that ^ ~ ^ = a7 ~ J/'

SOLUTION Let (xv yT) be the point of intersection of the curves

Then,

ax2 + by\ = I

b-a1 (b-a

a' x\ + b' y\ = 1

Differentiating (i) w.r.t. x, we get

2 ax + 2 by ^ = 0

...(iii)

...(iv)

ii - _ ax

= ± V ^ T a n d y = ± V ab2 dx

a + b """ 3 ' ' a + b

Thus, the two curves intersect at four points given by

by

(&) dx v y

axj

tyl •(v)

air a+b ' a+b_

Differentiating (i) and (ii) w.r.t. x, we get

dx a y

vlngfe of intersection at P

y

Differentiating (ii) w.r.t. x, we obtain

2a'x + 2b'y^ = 0

dy _ a' x dx b'y

AJ a2b ^ ab2 ] a + b ' a + b J'

dy dx

(xv yj

a Xi

b'y l ...(vi)


...(vii)

...(viii)

The two curves will intersect orthogonally, if

m-lm2 = -1 flXj a' Xj

aa' x\ = -bb'y2

Subtracting (iv) from (iii), we obtain

(a - a') x{ = - (b - b') y{

Dividing (viii) by (vii), we get

a-a' = b-b' aa' ~ bb'

I _ I = 1 _ 1 a b a' b'

x2 2

EXAMPLE 5 If the curves ~ + = 1 and y3 = 16x intersect at right

2 4 * angles, show that a = -. SOLUTION Let (Xj i/j) be the point of intersection of the two curves,

x2 2 Equations of two curves are ~ + = 1 and y3 = 16 x

Differentiating these w.r.t. x, we get

% + ^ ? = 0 a n d 3 y 2 ? = 16 2 4 dx dx a / j \

dx = — c, a2y

and dx 16 3y2

Let mj and m2 be the slopes of the tangents to the two curves at their point of intersection. Then,

•m — —J-dx _ v yc,

at (Xpyj)

=

V

16 3 y

If the two curves intersect at right angles, then

mj m2

"4*1 16 x — 2 « l/i 3yi

m l m 2 = - 1

= - 1

=> 64 Xj = 3 a2 y 3

=> 64 Xj = 3o2 x 16Xf [•_• (xv yj) lies on y3 = 16 x :. y3 = 16 x^

=> 4 = 3a2

=> a2 = 4/3. EXAMPLE 6 Show that the angle between the tangent at any point P and the line joining P to the origin O is the same at all points on the curve

SOLUTION We have,

logx2 + j/2) = k tan

Fig. 6.5


1 x2 ' "2

+y -4 X

x-f--y dx -

2 x + y dx

2x + ky = (far - 2y)

Ml dx)

,4 / dx

dy _ 2x + ky dx kx - 2y

Let the coordinates of P be (x]7 yj). Then,

dy' _ 2x1 + ky1

kxl - 2yx dx v J

If the tangent at P makes an angle 9 with x-axis, then

2 xl+kyx tan 0 = — kxx - 2yx

Suppose OP makes an angle 0 with x-axis. Then,


tan 0 = Slope of OP = y\ n

Let a be the angle between OP and PT. Then,

0 = a + 0

a = 0 - (J)

tan a = tan (0 - <f>)

tan 0 - tan (t> tan a 1 + tan 0 tan <

tan a =

tan a =

2x1 + ky1 y1

2x, + kyl y, 1 +-j— r—- X — kxx - 2y1 x1

2x12 + fcx1 y1 - kx-y yx+2y2

= 2 k x \ ~ 2 * a yi + 2x1y1 + ky1

2 *

a = tan - l k v v

= constant.

EXERCISE 6.2

1. Find the angle of intersection of the following curves :

(i) x2 + y2 = <2a2 and x 2 - y 2 = a2

(ii) x2 = y and x2 = 4 - y

(iii) y2 = 2x and x2 = 16y

(iv) x +y = 2a and xy = a

(v) 2x2 + y2 = 4 and y2 = 16x

(vi) x2 + y2 = 8ax and y2 (4a - x) = x3

2. Show that the curves -a + Xi b + X\

= 1

and, a2 + A2 b2 + X2

= 1

intersect orthogonally.

3. Find the condition so that the curves ^ + ̂ = 1 and 8. Find the angle of intersection of curves

4. Show that the curves x3 - 3xy2 = - 2 and 3x2y - y3 = 2 cut orthogonally.

5. Find the acute angles between the curves

y = 12x2 - 41 and y = j x2 - 51.

6. Show that the curves y2 = 4ax and ay2 = 4x3 intersect each other at an angle of tan"1 — and also if PGj and PG2 be the

normals to two curves at common point of intersection (other than the origin) meeting the axis of X in Gj and G2, then Ga G2 = 4a.

7. Prove that the curves y = e~ax and y = e~ax sin bx touch at , 1 . . (4n + l)n the points x = - — ~ — •

a2 b2

x2

2 , 2 = 1 may intersect orthogonally. y = [ | sin x | + | cos x | ] and x + y =5, where [•] denotes

the greatest integer function.

ANSWERS

1. « e =

(iv) 0

(ii) ± tan 1 (iii) ^ and ± tan 1 ^

(v) 71 (vi) tan 1 V2~

3. a2-b2 = A2- B2

8. t an - 1 2

5. ± tan -1 ( & § ) 23 v y

HINTS TO SELECTED PROBLEMS

4. Let (Xj,}/]) be the point of intersection of the two curves. Then,

dy dx

r j . . \

2 2 xi ~V\ 2*i Vi

and dx v y

= -^iVi 2 2'

c2 -Vi

Clearly, m J&L dx v y C,

dx , _ v = - 1

7. Show that the values o f o b t a i n e d from the two curves at dx

x = (4 n +1) 71 2b are equal.

8. We know that 1 < | sin x | + | cos x | < V2~for all x e R

:. y = [ | sin x | + | cos x | ] = 1

Thus, the two curves are y = 1 and x2 + y2 = 5 — • • • —


6.5 LENGTHS OF TANGENT, NORMAL, SUBTANGENT AND SUBNORMAL

Let the tangent and normal at a point P (x, y) on the curve y =f (x), meet the x-axis at T and N respectively. If G is the foot of the ordinate at P, then TG and GN are called the cartesian subtangent and subnormal, while the lengths PT and PN are called the lengths of the tangent and normal respectively.

If PT makes angle \j/ with x-axis, then tan =

we find that

Subtangent = TG = y cot \j/ =

dx . From Fig.6.6,

dy dx

dy -yfx Subnormal = GN = y tan \|/ •

Length of the tangent = PT = y cosec V|/

y' = v V l + cot \|/

Y

9 0 - Y \ ^P(X.Y)

/x* Y \

0 T G N »

Fig. 6.6

dx

dy dx

Length of the normal = PN = y sec y

= y \ 1 + tan2 y

y dx v /


EXAMPLE l Find the lengths of tangent, subtangent, normal and subnormal toy1 = 4ax at (at2, lat). SOLUTION We have,

y2 = 4ax

,*L =

dx dy _ 2a dx ~ y

= — = i ~ 2at t'

2y-?~ = 4A

dx v Now,

at , 2at)

Length of the tangent at (at , 2at) dx

dy dx

2at Vl + l/t2

1 It

2 Vt2 + 1 = 2 at t

2 at V f 2 + 1

dx Length of the normal at (at , 2at) = y

= 2 « t V l + i r

= 2a V f 2 + 1

? v 2at 9 Length of the subtangent at (atz, 2at) = = j j j - 2«r

Length of the subnormal at (at2 , 2at) = 2a fxy = 2a.

EXAMPLE 2 Prove that mth power of subtangent at any point on the curve xm + n = am'n y2" varies as the nth power of subnormal at the same point. SOLUTION We have,

xm + " = am~n y2"

n) logx = (m - n) log

m + n 0 + 2 n d y X y dx

dy = m + n dx 2 n X V /

/ j = length of the subtangent at (x, y)

y 2nx dy (m + n) dx

and,

l2 = length of the subnormal at (x, y)

= y dy _ (m + n)y dx 2 nx


Now,

and,

' 2 nx (m + ri)

yn = (2n) mx m

(m + n)m

hn = (m + n)y 2nx

= (m + nf y2" (2k)"

= (2n)m + "xm + " Z2" (m + n)m + V "

_ (2 n)m + n

• (m + n)m + " = constant

Hence, l\m varies as l2n.

a ,m-n 2n y

EXERCISE 6.3

If the subnormal of the curve xyn = an+ is of constant length, prove that n = - 2 .

x2 v2

2. For the ellipse ~ + , = 1/ prove that the length of the it b1

normal at any point varies inversely as perpendicular from origin on any tangent at the same point.

3. For the curve xmy" = am + prove that the subtangent at any point varies as the abscissa of the point. 4. Find the lengths of tangent and normal at point t = rc/3 on the curve x = a (t + sin t), y = a (1 - cos t). 5. Prove that for the curve y = a log (x -a), sum of the tan-gent and subtangent varies as the product of the coordinates of the point of contact.

ANSWERS

4. a, a!A3

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Tangents & Normal RDSA