#### Provide solution for RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 30 maths textbook solution.

Answer : (a) is the correct option

Hint : Equation of the normal,  $y-y_{1}=N\left(x-x_{1}\right)$

Given : $3 y=6 x-5 x^{3}$

Solution :

Let  $\left(x_{1}, y_{1}\right)$ be the point on the curve $3 y=6 x-5 x^{3}$ at which normal passes through origin

$3 y=6 x-5 x^{3}$                                                          ....(1)

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

\begin{aligned} &3 \frac{d y}{d x}=6-15 x^{2} \\ &\frac{d y}{d x}=2-5 x^{2} \end{aligned}

Slope of the tangent at P is $2-5x_{1}^{2}$

Slope of normal at P is $\frac{1}{5x{_{1}}^{2}-2}$

Equation of normal at P is

$y-y_{1}=\frac{1}{5 x_{1}^{2}-2}\left(x-x_{1}\right)$

Since it passes through the origin

$y=\frac{x}{5 x_{1}^{2}-2}$                                                          .....(2)

On solving eqn 1 and 2

We get $\mathrm{x}=1 \text { and } \mathrm{y}=\frac{1}{3}$

As we know, abscissa is the x-coordinate

So, option (a) satisfies above equation i.e., 1