#### Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 15 maths Textbook Solution.

Answer: The requried point are $(0,0) \&(2 a,-2 a)$

Hint:$\text { The slope of the tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

Given:The Curve is $2 a^{2} y=x^{3}-3 a x^{2} \rightarrow(1)$

Solution:$2 a^{2} y=x^{3}-3 a x^{2}$

$\text { Differentiating the above with respect to } x$

$\therefore \frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0$

$2 a^{2} \frac{d y}{d x}=3 x^{3-1}-3 a(2)(x)^{2-1}$

$2 a^{2} \frac{d y}{d x}=3 x^{2}-6 a x$

$\frac{d y}{d x}=\frac{1}{2 a^{2}}\left[3 x^{2}-6 a x\right]$

$\text { Slope } \mathrm{m}_{1}=\frac{d y}{d x}=\frac{1}{2 a^{2}}\left[3 x^{2}-6 a x\right] \rightarrow(2)$

Also

$\text { Slope } \mathrm{m}_{2}=\frac{d y}{d x}=\tan \theta$

$\tan 0^{\circ}=0$

$\text { [slope is parallel to } x \text { -axis] }$

$m_{1}=m_{2}$

$\frac{1}{2 a^{2}}\left[3 x^{2}-6 a x\right]=0$

$3 x^{2}-6 a x=0$

\begin{aligned} &3 x(x-2 a)=0 \\ &3 x=0 \text { or } x-2 a=0 \\ &x=0 \text { or } x=2 a \end{aligned}

$\text { When put } x=0 \text { in eqn(1) } 2 a^{2} y=x^{3}-3 a x^{2}$

\begin{aligned} &2 a^{2} y=(0)^{3}-3 a(0)^{2} \\ &2 a^{2} y=0 \\ &y=0 \end{aligned}

$\text { When put } x=2 a \text { in eqn (1) } 2 a^{2} y=x^{3}-3 a x^{2}$

\begin{aligned} &2 a^{2} y=(2 a)^{3}-3 a(2 a)^{2} \\ &2 a^{2} y=8 a^{3}-3 a\left(4 a^{2}\right) \\ &2 a^{2} y=8 a^{3}-12 a^{3} \\ &2 a^{2} y=-4 a^{3} \\ &y=-2 a \end{aligned}

Thus the requried points are $(0,0) \&(2 a,-2 a)$