#### Provide Solution For R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 17 Sub Question 2 Maths Textbook Solution.

Answer:$\text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(2,0) \&(-2,0)$

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

Given: $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1 \quad \rightarrow(1)$

Solution: $\text { Differentiating eqn(1) with respect to } x$

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

$\frac{2 x^{2-1}}{4}+\frac{2 y}{25} \frac{d y}{d x}=0$

$\frac{2 x}{4}+\frac{2 y}{25} \frac{d y}{d x}=0$

\begin{aligned} &\frac{x}{2}+\frac{2 y}{25} \frac{d y}{d x}=0 \\ &\frac{2 y}{25} \frac{d y}{d x}=\frac{-x}{2} \\ &\frac{d y}{d x}=\frac{-x}{2} \times \frac{25}{2 y} \\ &\frac{d y}{d x}=\frac{-25 x}{4 y} \& \frac{d x}{d y}=\frac{-4 y}{25 x} \end{aligned}

Now the tangent is parallel to the y-axis if the slope of the normal is zero
Slope of normal=$\frac{-1}{\text { Slope of tangent }} \Rightarrow \frac{-1}{\frac{-25 x}{4 y}}=0$

$\frac{4 y}{25 x}=0$

$\text { This is possible if } y=0$

$\text { Then, } \frac{x^{2}}{4}+\frac{y^{2}}{25}=1 \text { for } y=0$

$\frac{x^{2}}{4}+\frac{0}{25}=1$

$x^{2}=4$

$x=\pm 2$

Thus $\text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(2,0) \&(-2,0)$