Question

Prove that the curves x = y 2 and xy = k cut at right angles if 8 k 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

Equations of the given curves are,

x= y 2 xy=k

From the above equations,

y 3 =k y= k 2 3 x= k 1 3

Hence, the intersection point of the two curves is ( k 2 3 , k 1 3 ).

To find slope of the curve x= y 2 , we need to differentiate the equation with respect to x.

1=2y( dy dx ) dy dx = 1 2y

Therefore, slope of the tangent to the curve x= y 2 at ( k 2 3 , k 1 3 ) is,

( dy dx ) ( k 2 3 , k 1 3 ) = 1 2 k 1 3

To find slope of the curve xy=k, we need to differentiate the equation with respect to x.

x( dy dx )+y=0 dy dx = −y x

Therefore, slope of the tangent to the curve xy=k at ( k 2 3 , k 1 3 ) is,

( dy dx ) ( k 2 3 , k 1 3 ) =− k 1 3 k 2 3 =− 1 k 1 3

When two curves intersect with each other at right angles at ( k 2 3 , k 1 3 ), the product of their tangents must be −1; therefore,

( 1 2 k 1 3 )( −1 k 1 3 )=−1 2 k 2 3 =1 ( 2 k 2 3 ) 3 = ( 1 ) 3 8 k 2 =1

Hence, it is proved that two curves cut each other at right angles when 8 k 2 =1.