Question

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

Answer 1

Hint: To solve this question, we will find out the intersection point, i.e. where the two curves cut each other in terms of k. Then we will find the slope of both the curves $x={{y}^{2}}$ and xy = k at the intersection point. Then, we will equate their product to – 1 because the tangents are perpendicular at that point, to get the value of k.

Complete step-by-step solution -
To start with, we will find the intersection point P of these curves in terms of k. The two curves given are,
$x={{y}^{2}}.....\left( i \right)$
$xy=k......\left( ii \right)$
Now, we will put the value of $x={{y}^{2}}$ from equation (i) to equation (ii). Thus, we will get the following result.
$\Rightarrow {{y}^{2}}.y=k$
$\Rightarrow {{y}^{3}}=k$
$\Rightarrow y={{k}^{\dfrac{1}{3}}}....\left( iii \right)$
Now, we will put the value of y from equation (iii) to equation (i). Thus, we will get,
$\Rightarrow x={{\left( {{k}^{\dfrac{1}{3}}} \right)}^{2}}$
$\Rightarrow x={{k}^{\dfrac{2}{3}}}.....\left( iv \right)$
Thus, the intersection point of the curves is $P\left( {{k}^{\dfrac{2}{3}}},{{k}^{\dfrac{1}{3}}} \right).$ Now, it is given that the curves cut each other at right angles at the intersection point P. So, the tangents of these curves at P will also be perpendicular to each other. So, we will find the slope of both the curves at $P\left( {{k}^{\dfrac{2}{3}}},{{k}^{\dfrac{1}{3}}} \right).$
Slope of $x={{y}^{2}}:$
$x={{y}^{2}}$
$\Rightarrow y=\sqrt{x}$
On differentiating both the sides with respect to x, we get,
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{x} \right)$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\times \dfrac{1}{\sqrt{x}}$
At any point P, let us take the slope as,
${{\left( \dfrac{dy}{dx} \right)}_{{{P}_{1}}}}=\dfrac{1}{2}\times \dfrac{1}{\sqrt{{{k}^{\dfrac{2}{3}}}}}$
${{\left( \dfrac{dy}{dx} \right)}_{{{P}_{1}}}}=\dfrac{1}{2{{k}^{\dfrac{1}{3}}}}.....\left( v \right)$
Slope of xy = k:
$\Rightarrow y=\dfrac{k}{x}$
On differentiating both the sides with respect to x, we get,
$\dfrac{dy}{dx}=k.\dfrac{d}{dx}\left( \dfrac{1}{x} \right)$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-k}{{{x}^{2}}}$
At any point P, let us take the slope as,
${{\left( \dfrac{dy}{dx} \right)}_{{{P}_{2}}}}=\dfrac{-k}{{{\left( {{k}^{\dfrac{2}{3}}} \right)}^{2}}}$
$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{{{P}_{2}}}}=\dfrac{-k}{{{k}^{\dfrac{4}{3}}}}$
$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{{{P}_{2}}}}=\dfrac{-1}{{{k}^{\dfrac{1}{3}}}}....\left( vi \right)$
Now, these tangents are perpendicular to each other. If two lines with slope ${{m}_{1}}$ and ${{m}_{2}}$ are perpendicular to each other, then we have,
${{m}_{1}}{{m}_{2}}=-1$
Similarly, for our case, we can say that,
${{\left( \dfrac{dy}{dx} \right)}_{{{P}_{1}}}}.{{\left( \dfrac{dy}{dx} \right)}_{{{P}_{2}}}}=-1.....\left( viii \right)$
From (v), (vi) and (vii), we will get,
$\left( \dfrac{1}{2{{k}^{\dfrac{1}{3}}}} \right)\left( \dfrac{-1}{{{k}^{\dfrac{1}{3}}}} \right)=-1$
$\Rightarrow \dfrac{1}{2{{k}^{\dfrac{1}{3}}}\times {{k}^{\dfrac{1}{3}}}}=1$
$\Rightarrow {{k}^{\dfrac{1}{3}}}\times {{k}^{\dfrac{1}{3}}}=\dfrac{1}{2}$
$\Rightarrow {{k}^{\dfrac{2}{3}}}=\dfrac{1}{2}$
On cubing both the sides, we will get,
${{\left( {{k}^{\dfrac{2}{3}}} \right)}^{3}}={{\left( \dfrac{1}{2} \right)}^{3}}$
$\Rightarrow {{k}^{\dfrac{2}{3}\times 3}}=\dfrac{1}{{{2}^{3}}}$
$\Rightarrow {{k}^{2}}=\dfrac{1}{8}$
$\Rightarrow 8{{k}^{2}}=1$.

Note: After finding out the intersection point, we can also approach the solution in the following way:
Slope of $x={{y}^{2}}:$
$\dfrac{d}{dx}\left( x \right)=\dfrac{d}{dx}\left( {{y}^{2}} \right)$
$\Rightarrow 1=2y\dfrac{dy}{dx}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2y}$
${{\left( \dfrac{dy}{dx} \right)}_{{{P}_{1}}}}=\dfrac{1}{2{{k}^{\dfrac{1}{3}}}}$
Slope of xy = k
$\dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( k \right)$
$\Rightarrow x\dfrac{dy}{dx}+y=0$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-y}{x}$
${{\left( \dfrac{dy}{dx} \right)}_{{{P}_{2}}}}=\dfrac{-{{k}^{\dfrac{1}{3}}}}{{{k}^{\dfrac{2}{3}}}}$
${{\left( \dfrac{dy}{dx} \right)}_{{{P}_{2}}}}=\dfrac{-1}{{{k}^{\dfrac{1}{3}}}}$
Now, the tangents are perpendicular, so,
$\dfrac{1}{2{{k}^{\dfrac{1}{3}}}}\times \left( \dfrac{-1}{{{k}^{\dfrac{1}{3}}}} \right)=-1$
$\Rightarrow {{k}^{\dfrac{2}{3}}}=\dfrac{1}{2}$
$\Rightarrow {{k}^{2}}=\dfrac{1}{8}$
$\Rightarrow 8{{k}^{2}}=1$.