Question

prove that the curves x=y² and xy=k , cut at right angle if 8k²=1 ​

Answer 1

AnswEr:-

HELLO DEAR,

Given curves are x = y²-------( 1 ) ,

xy = k -----------( 2 )

=> x = k/y

x = y² => k/y = y²

k = y³

y = k⅓ [ put in ( 2 )

x = k/y

x = k/k⅓

hence, the point of intersection of of curves are (k⅔ , k⅓)

NOW,

we know slope of tangent = dy/dx .

for x = y²

2y*dy/dx = 1

dy/dx = 1/2y-------( 3 )

and slope of tangent at (k⅔ , k⅓) is

$\frac{dy}{dx} k^{ \frac{2}{3} } , {k}^{ \frac{1}{3} } = \frac{1}{2( {k}^{ \frac{1}{3} }) } = \frac{1}{2 {k}^{ \frac{1}{3} } }$

for xy = k

x*dy/dx + y = 0

dy/dx = -y/x

the slope of tangent at (k⅔ , k⅓) is

$\frac{dy}{dx} {k}^{ \frac{2}{3} } , {k}^{ \frac{1}{3} } = \frac{ - {k}^{ \frac{1}{3} } }{ {k}^{ \frac{2}{3} } } = \frac{ - 1}{ {k}^{ \frac{1}{3} } }$

we know if two lines are perpendicular then product of there slope = -1

now,

(slope of tangent at curve x = y²) × (slope of tangent at curve xy = k) = -1

$\frac{1}{2 {k}^{ \frac{1}{3} } } = \frac{ - 1}{ {k}^{ \frac{1}{3} } } = - 1 \\ \\ \frac{1}{2 {k}^{ \frac{1}{3} + \frac{1}{3} } } = 1 \\ \\ 1 = 2 {k}^{ \frac{2}{3} }$

on cubing both sides, we get,

$1 = 8 {k}^{2} \\ \\ hence \: \boxed{8 {k}^{2} = 1 }$

#ANSWERWITHQUALITY

#BAL

Answer 2

Answer:

Step-by-step explanation: