Question

If the normal to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ makes an angle $\phi$ with the $x-$ axis, then its equation is

• A. $x\sin \left(\phi \right)+y\cos \left(\phi \right)=a$
• B. $y\cos \left(\phi \right)-x\sin \left(\phi \right)=a\cos \left(2\phi \right)$
• C. $x\cos \left(\phi \right)+y\sin \left(\phi \right)=a\sin \left(2\phi \right)$
• D. None of these

Answer 1

The correct option is B

$y\cos \left(\phi \right)-x\sin \left(\phi \right)=a\cos \left(2\phi \right)$

Step 1: Determine the slope of the normal to the curve

The given equation of the curve: $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)=\frac{d}{dx}\left(a^{\frac{2}{3}}\right) \\ \Rightarrow \frac{2}{3}{x}^{-\frac{1}{3}}+\frac{2}{3}{y}^{-\frac{1}{3}}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{x^{-{\displaystyle \frac{1}{3}}}}{y^{-\frac{1}{3}}} \\ \Rightarrow \frac{dy}{dx}=-{\left(\frac{y}{x}\right)}^{\frac{1}{3}} \\ \Rightarrow -\frac{dx}{dy}={\left(\frac{x}{y}\right)}^{\frac{1}{3}} \end{gathered}$$

Thus, the slope of the normal to the curve at the point $\left(h,k\right)$ can be given by ${\left(-\frac{dx}{dy}\right)}_{\left(h,k\right)}={\left(\frac{h}{k}\right)}^{\frac{1}{3}}$.

Step 2: Determine the value of $h$ and $k$

It is given that the normal makes an angle $\phi$ with the $x-$ axis.

Thus, $\tan \left(\phi \right)={\left(\frac{h}{k}\right)}^{\frac{1}{3}}$.

$$\begin{gathered} \Rightarrow \frac{\sin \left(\phi \right)}{\cos \left(\phi \right)}={\left(\frac{h}{k}\right)}^{\frac{1}{3}} \\ \Rightarrow \frac{{\sin }^2\left(\phi \right)}{{\cos }^2\left(\phi \right)}={\left(\frac{h}{k}\right)}^{\frac{2}{3}} \\ \Rightarrow \frac{h^{{\displaystyle \frac{2}{3}}}}{{\sin }^2\left(\phi \right)}=\frac{k^{{\displaystyle \frac{2}{3}}}}{{\cos }^2\left(\phi \right)} \\ \Rightarrow \frac{h^{{\displaystyle \frac{2}{3}}}}{{\sin }^2\left(\phi \right)}=\frac{k^{{\displaystyle \frac{2}{3}}}}{{\cos }^2\left(\phi \right)}=\frac{h^{{\displaystyle \frac{2}{3}}}+k^{{\displaystyle \frac{2}{3}}}}{{\sin }^2\left(\phi \right)+{\cos }^2\left(\phi \right)} \\ \Rightarrow \frac{h^{{\displaystyle \frac{2}{3}}}}{{\sin }^2\left(\phi \right)}=\frac{k^{{\displaystyle \frac{2}{3}}}}{{\cos }^2\left(\phi \right)}=\frac{h^{{\displaystyle \frac{2}{3}}}+k^{{\displaystyle \frac{2}{3}}}}{1}\left[\because {\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1\right] \end{gathered}$$

As the equation of the curve is $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ and the point $\left(h,k\right)$ is on the curve, thus $h^{\frac{2}{3}}+k^{\frac{2}{3}}=a^{\frac{2}{3}}$.

So, $\Rightarrow \frac{h^{{\displaystyle \frac{2}{3}}}}{{\sin }^2\left(\phi \right)}=\frac{k^{{\displaystyle \frac{2}{3}}}}{{\cos }^2\left(\phi \right)}=a^{\frac{2}{3}}$.

Therefore, $\frac{h^{{\displaystyle \frac{2}{3}}}}{{\sin }^2\left(\phi \right)}=a^{\frac{2}{3}}$.

$\Rightarrow h=a{\sin }^3\left(\phi \right)$.

Therefore, $\frac{k^{{\displaystyle \frac{2}{3}}}}{{\cos }^2\left(\phi \right)}=a^{\frac{2}{3}}$.

$\Rightarrow k=a{\cos }^3\left(\phi \right)$.

Step 3: Determine the equation of the normal

The equation of the normal can be given by $\left(y-k\right)=\tan \left(\phi \right)\left(x-h\right)$.

$$\begin{gathered} \Rightarrow \left(y-a{\cos }^3\phi \right)=\tan \left(\phi \right)\left(x-a{\sin }^3\phi \right) \\ \Rightarrow \left(y-a{\cos }^3\phi \right)=\frac{\sin \left(\phi \right)}{\cos \left(\phi \right)}\left(x-a{\sin }^3\phi \right) \\ \Rightarrow y\cos \left(\phi \right)-a{\cos }^4\left(\phi \right)=x\sin \left(\phi \right)-a{\sin }^4\left(\phi \right) \\ \Rightarrow y\cos \left(\phi \right)-x\sin \left(\phi \right)=a{\cos }^4\left(\phi \right)-a{\sin }^4\left(\phi \right) \\ \Rightarrow y\cos \left(\phi \right)-x\sin \left(\phi \right)=a\left[{\cos }^4\left(\phi \right)-{\sin }^4\left(\phi \right)\right] \\ \Rightarrow y\cos \left(\phi \right)-x\sin \left(\phi \right)=a\left[\left\{{\cos }^2\left(\phi \right)-{\sin }^2\left(\phi \right)\right\}\left\{{\cos }^2\left(\phi \right)+{\sin }^2\left(\phi \right)\right\}\right] \\ \Rightarrow y\cos \left(\phi \right)-x\sin \left(\phi \right)=a\left[\cos \left(2\phi \right)\times 1\right]\left[\because {\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)=\cos \left(2\theta \right)\right] \\ \Rightarrow y\cos \left(\phi \right)-x\sin \left(\phi \right)=a\cos \left(2\phi \right) \end{gathered}$$

Therefore, the equation of the normal $y\cos \left(\phi \right)-x\sin \left(\phi \right)=a\cos \left(2\phi \right)$.

Hence, (B) is the correct option.