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

Answers (1)

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


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