Question

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.]

Answer 1

The equations of the given curves are given as Putting x = y² in xy = k, we get:

y³=k=y=k\(^{\frac{1}{3}}\)

=x=k\(^{\frac{2}{3}}\)

Thus, the point of intersection of the given curves is (k\(^{\frac{2}{3}}\),k\(^{\frac{1}{3}}\)).

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

1=2y \(\frac{dy}{dx}\)=\(\frac{dy}{dx}\)=\(\frac{1}{2}\)y

Therefore, the slope of the tangent to the curve x = y2 at (k\(^{\frac{2}{3}}\).) \(\frac{dy}{dx}\)]_(k\(^{\frac{2}{3}}\)_,k\(^{\frac{3}{3}}\)₎=\(\frac{1}{2k^{\frac{1}{3}}}\)

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

x \(\frac{dy}{dx}\)+y=0=\(\frac{dy}{dx}\)=\(-\frac{y}{x}\)

Slope of the tangent to the curve xy = k at (\(\frac{2}{k^3}\),\(\frac{1}{k^3}\)) is given by,

\(\frac{dy}{dx}\)](k\(^{\frac{2}{3}}\),\({\frac{1}{k^3}}\))=\(-\frac{y}{x}\)](\(\frac{2}{k^3}\),\(\frac{3}{k^3}\))=\(\frac{\frac{-k_1}{3}}{\frac{K_2}{3}}\)=\(\frac{\frac{-1}{k_1}}{3}\)

We know that two curves intersect at right angles if the tangents to the curves at the

point of intersection i.e., at(\(\frac{2}{k^3}\),\(\frac{1}{k^3}\)) 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 (\(\frac{2}{k^3}\),\(\frac{1}{k^3}\)) is −1.

=\(\frac{2k^2}{3}\)=1

=\((\frac{2k^2}{3})^3\)=(1)3

=8k²=1

Hence, the given two curves cut at right angles if 8k² = 1.