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.