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

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- $\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.

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