Question

The points on the curve $9y^2=x^3$ where the normal to the curve makes equal intercepts with coordinates axes is :

• A. $(4,\frac{8}{3})or(4,-\frac{8}{3})$
• B. $(-4,\frac{8}{3})$
• C. $(-4,-\frac{8}{3})$
• D. None of these

Answer 1

The correct option is A $(4,\frac{8}{3})or(4,-\frac{8}{3})$

Let the point on the curve $9y^2=x^3$ be $P(x_1,y_1)$
differentiating the curve w.r.t $x$
$2\times 9\times y\times \frac{dy}{dx}=3x^2$
$\Rightarrow \frac{dy}{dx}=\frac{x^2}{2\times 3y}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{x_1^2}{2\times 3y_1}$
Thus slope of Normal at P $=-\frac{dy}{dx}=\frac{-3y_1\times 2}{x_1^2}$
Given normal makes equal intercept
$\Rightarrow \frac{-3y_1\times 2}{x_1^2}=\pm 1$
$\Rightarrow \frac{-3y_1\times 2}{x_1^2}=1$
$\Rightarrow -3y_1\times 2=x_1^2$ so $9y_1^2\times 4=x_1^4\Rightarrow x_1^3\times 4=x_1^4$
So $x_1=4,y=-\frac{8}{3}$
$\Rightarrow \frac{-3y_1\times 2}{x_1^2}=-1$
$\Rightarrow 2\times 3y_1=x_1^2$ ..........(1)
Also point lies point on the curve $9y^2=x^3$
$\Rightarrow 9y_1^2=x_1^3$
$\Rightarrow 9\times \frac{x_1^4}{36}=x_1^3$
$\Rightarrow x_1=4$
$\Rightarrow x_1=4,y_1=\frac{16}{6}=\frac{8}{3}$
$\Rightarrow (4,\frac{8}{3})$
Finally the points are $(4,\frac{8}{3})or(4,\frac{-8}{3})$
Hence, option 'A' is correct.