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

The equations of the given curves are given as

Putting x = y² in xy = k, we get:

x = 1, x = -2 or x = 3

x = 1, x = -2 and x = 3

Thus, the point of intersection of the given curves is (-∞, -2), (-2, 1), (1, 3).

Differentiating x = y² with respect to x, we have:

(3, ∞)

Therefore, the slope of the tangent to the curve x = y² at (-∞, -2) is (-2, 1) 

On differentiating xy = k with respect to x, we have:

(1, 3) 

∴ Slope of the tangent to the curve xy = k at(3, ∞)is given by,

(-∞, -1), (-1, 1) and (1, ∞)

The two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at(-∞, -1) 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 (-1, 1) is −1.

i.e., (1, ∞)