Question

Prove that the curves x = y^ 2 and xy = k cut at right angles if 8k^ 2 = 1. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]

Answer 1

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

for xy = k

x*dy/dx + y = 0

dy/dx = -y/x

the slope of tangent at (k⅔ , k⅓) is

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

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

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

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

$\bold{1 = 2k^{2/3}}$

on cubing both side,

$\bold{1 = 8k^2}$

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

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

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

\bold{\frac{dy}{dx}_{k^{2/3} , k^{1/3}} = \frac{1}{2(k^{1/3})} = 1/2k^{1/3}}dxdyk2/3,k1/3=2(k1/3)1=1/2k1/3

for xy = k

x*dy/dx + y = 0

dy/dx = -y/x

the slope of tangent at (k⅔ , k⅓) is

\bold{\frac{dy}{dx}_{k^{2/3} , k^{1/3}} = \frac{-k^{1/3}}{k^{2/3}} = \frac{-1}{k^{1/3}}}dxdyk2/3,k1/3=k2/3−k1/3=k1/3−1

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

\bold{\frac{1}{2k^{1/3}} = \frac{-1}{k^{1/3}} = -1}2k1/31=k1/3−1=−1

\bold{\frac{1}{2k^{1/3 + 1/3}} = 1}2k1/3+1/31=1

\bold{1 = 2k^{2/3}}1=2k2/3

on cubing both side,

\bold{1 = 8k^2}1=8k2

hence, \bold{\boxed{8k^2 = 1}}8k2=1

I HOPE ITS HELP YOU DEAR,

THANKS