Question

Prove that the curves $x=y^2$ and xy=k cut at right angle if $8k^2=1$.

Answer 1

The equation of the given curves are $x=y^2$ .....(i)

and xy=k .......(ii)

The two curves meet where $\frac{k}{y}=y^2$[Eliminating x between Eqs. (i) and (ii)]

$\Rightarrow y^3=k\Rightarrow y=k^{1/3}$

Substituting this value of y in Eq. (i), we get $x=(k^{1/3}{)}^2=k^{2/3}$

$\therefore$ Eq. (i) and Eq. (ii) intersect at the point $=\frac{1}{2k^{1/3}}$

On differentiating Eq. (i), w.r.t. x, we get $1=2y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$

$\therefore$ Slope of the tangent to the first curve Eq. (i) at $(k^{2/3},k^{1/3})$

$=\frac{1}{2k^{1/3}}$ ...(iii)

From Eq. (ii), $y=\frac{k}{x}\Rightarrow \frac{dy}{dx}=-\frac{k}{x^2}$

$\therefore$ Slope of the tangent to the second curve Eq. (ii) at $(k^{2/3},k^{1/3})$

$=-\frac{k}{(k^{2/3}{)}^2}=-\frac{1}{k^{1/3}}$ ...(iv)

We know that two curves intersect at right angles, if the tangents to the curves at the point of intersection i.e.,at -- are perpendicular to each other.

This implies that we should have the product of the tangents =-1

$\left(\frac{1}{2k^{1/3}}\right)(-\frac{1}{k^{1/3}})=-1\Rightarrow 1=2k^{2/3}\Rightarrow 1^3=(2k^{2/3}{)}^3\Rightarrow 1=8k^2$

Hence, the given two curves cut at right angles, if $8k^2=1$. Hence, proved.