#### Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 25 maths textbook solution.

Answer : (a) is the correct option

Hint : Slope of the mormal is 1 or -1

Given :  $9y^{2}=x^{3}$

Solution :

$9y^{2}=x^{3}$                                     (1)

Differentiating (1) w.r.t x, we get

\begin{aligned} &18 y \frac{d y}{d x}=3 x^{2} \\ &\frac{d y}{d x}=\frac{3 x^{2}}{18 y}=\frac{x^{2}}{6 y} \end{aligned}

\begin{aligned} &\text { Slope of normal }=-\frac{d x}{d y}=-\frac{6 y}{x^{2}}=1 \text { or }-1 \\ &\text { Case } 1, \mathrm{~m}=1 \end{aligned}

\begin{aligned} &-\frac{6 y}{x^{2}}=1 \\ &-6 y=x^{2} \\ &-6 y_{1}=x_{1}^{2} \\ &y_{1}=-\frac{x_{1}^{2}}{6} \end{aligned}

\begin{aligned} &P\left(x, \frac{x^{2}}{6}\right) \\ &\text { Putting } P\left(x, \frac{x^{2}}{6}\right) \ln (1) \end{aligned}

\begin{aligned} &9\left(\frac{x^{2}}{6}\right)^{2}=x^{3} \\ &\Rightarrow 9 \times \frac{x^{4}}{36}=x^{3} \\ &\Rightarrow x^{3}(x-4)=0 \end{aligned}

$\Rightarrow x= 0,4$

Putting the values of x in (1) we get $y=-\frac{8}{3}$

Similarly in case (2) if $y=\pm\frac{8}{3}$

$\therefore$ The point $\left(4, \pm \frac{8}{3}\right)$