NCERT Solutions
Class 12 Maths
Application of Derivatives
Ex.6.3 Q.23
Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1
[Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]
The equations of the given curves are given as x = y2 and xy = k
Putting x = y2 in xy = k, we get:
y3 = k
⟹ y = k1/3
So, x = k2/3
Thus, the point of intersection of the given curves is (k2/3, k1/3).
Differentiating x = y2 with respect to x, we have:
1 = 2y 
⟹ 
Therefore, the slope of the tangent to the curve x = y2 at (k2/3, k1/3) is
⟹ = 
On differentiating xy = k with respect to x, we have:
x + y = 0
⟹ = 
So, the slope of the tangent to the curve xy = k at (k2/3, k1/3) is
⟹ = 
= 
We know that two curves intersect at right angles if the tangents to the curves at the point of
intersection i.e., at (k2/3, k1/3) are perpendicular to each other.
This implies that we should have the product of the tangents as − 1.
Thus, the given two curves cut at right angles if the product of the slopes of their respective
tangents at is −1.
i.e. (
) ( ) = -1
⟹ 2k2/3 = 1
⟹ (
)3 = 13
⟹ 8k2 = 1
Hence, the given two curves cut at right angel if 8k2 = 1.