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# Choose the correct answer in the points on the curve $9y^2 = x^3$, where the normal to the curve makes equal intercepts with the axes are

$\begin{array}{l \qquad l \qquad l} (A)\; \left(4,\pm {\frac{8}{3}}\right) \quad &(B)\; \left(4,{\frac{-8}{3}}\right) \\ (C)\;\left(4,\pm {\frac{3}{8}}\right) \quad &(D)\;\left(\pm 4, {\frac{3}{8}}\right)\end{array}$

Toolbox:
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} Step 1: Given : 9y^2=x^3 Differentiating with respect to x 18y.\large\frac{dy}{dx}=$$3x^2$
$\large\frac{dy}{dx}=\frac{3x^2}{18y}$
Step 2:
Let $p(x_1,y_1)$ be the point where normal is drawn.
Slope of the tangent =$\large\frac{x_1^2}{6y_1^2}$
Slope of the normal =$\large\frac{-6y_1^2}{x_1^2}$
Normal make equal intercepts on the curve
Its slope =$\pm 1$
$\large\frac{-6y_1}{x_1^2}$$=\pm 1 6y_1=\pm x_1^2----(1) Step 3: (x_1,y_1) lies on the curve 9y^2=x^3 9y_1^2=x_1^3------(2) Taking +ve sign,eliminating y_1 from (1) & (2) 9.\big(\large\frac{x_1^2}{6}\big)^2=$$x_1^3$
$\large\frac{9x_1^4}{36}$$=x_1^3$
$x_1=4$
Step 4:
$y_1=\pm \large\frac{x_1^2}{6}$
by substituting the value of $x_1=4$ we get
$\quad=\pm\large\frac{16}{6}$
$\quad=\pm\large\frac{8}{3}$
The point $P$ is $(4,\pm\large\frac{8}{3})$
Part (A) is the correct answer.
edited Sep 2, 2013