The points on the curve


The points on the curve $9 y^{2}=x^{3}$, where the normal to the curve makes equal intercepts with the axes are

(A) $\left(4, \pm \frac{8}{3}\right)$

(B) $\left(4, \frac{-8}{3}\right)$

(C) $\left(4, \pm \frac{3}{8}\right)$

(D) $\left(\pm 4, \frac{3}{8}\right)$


The equation of the given curve is $9 y^{2}=x^{3}$.

Differentiating with respect to x, we have:

$9(2 y) \frac{d y}{d x}=3 x^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{x^{2}}{6 y}$

The slope of the normal to the given curve at point is

$y-y_{1}=\frac{-6 y_{1}}{x_{1}^{2}}\left(x-x_{1}\right)$

$\Rightarrow x_{1}^{2} y-x_{1}^{2} y_{1}=-6 x y_{1}+6 x_{1} y_{1}$

$\Rightarrow 6 x y_{1}+x_{1}^{2} y=6 x_{1} y_{1}+x_{1}^{2} y_{1}$

$\Rightarrow \frac{6 x y_{1}}{6 x_{1} y_{1}+x_{1}^{2} y_{1}}+\frac{x_{1}^{2} y}{6 x_{1} y_{1}+x_{1}^{2} y_{1}}=1$

$\Rightarrow \frac{x}{\frac{x_{1}\left(6+x_{1}\right)}{6}}+\frac{y}{\frac{y_{1}\left(6+x_{1}\right)}{x_{1}}}=1$

It is given that the normal makes equal intercepts with the axes.

Therefore, We have:

$\therefore \frac{x_{1}\left(6+x_{1}\right)}{6}=\frac{y_{1}\left(6+x_{1}\right)}{x_{1}}$

$\Rightarrow \frac{x_{1}}{6}=\frac{y_{1}}{x_{1}}$

$\Rightarrow x_{1}^{2}=6 y_{1}$    ....(1)

Also, the pointlies on the curve, so we have

$9 y_{1}^{2}=x_{1}^{3}$    ....(2)

From (i) and (ii), we have:

$9\left(\frac{x_{1}^{2}}{6}\right)^{2}=x_{1}^{3} \Rightarrow \frac{x_{1}^{4}}{4}=x_{1}^{3} \Rightarrow x_{1}=4$

From (ii), we have:

$9 y_{1}^{2}=(4)^{3}=64$

$\Rightarrow y_{1}^{2}=\frac{64}{9}$

$\Rightarrow y_{1}=\pm \frac{8}{3}$

Hence, the required points are $\left(4, \pm \frac{8}{3}\right)$.

The correct answer is A.





Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now