Question

Prove that the curves x = y2 and xy = k cut at right angles if
8k2 = 1.
​

Answer 1

Step-by-step explanation:

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

(dy/dx) k^{2/3} , k^{1/3} = 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

dy/dx{k^{2/3} , k^{1/3}} = -k^{1/3}/k^{2/3} = -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

1/2k^{1/3} = -1/k^{1/3} = -1

1/{2k^{1/3 + 1/3}} = 1

1 = 2k^{2/3}

on cubing both side,

1 = 8k^2

hence, 8k^2 = 1

I HOPE ITS HELP YOU

Answer 2

Answer:

Step-by-step explanation: