#### Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 19 Sub Question 2 Maths Textbook Solution.

Answer: $\text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(3,0) \&(-3,0)$

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

Given: $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \quad \rightarrow(1)$

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

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0 \\ &\frac{2 x}{9}+\frac{2 y}{16} \frac{d y}{d x}=0 \\ &\frac{d y}{d x}=\frac{-16}{9} \frac{x}{y} \end{aligned}

$\text { Now the tangent is parallel to the } \mathrm{y} \text { -axis if the slope of the normal is zero }$

$\text { slope of the normal }=\frac{-1}{\left(\frac{-16 x}{9 y}\right)} \Rightarrow \frac{9 y}{16 x}=0 \text { which is possible if } y=0$

Then,

$\frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \text { for } \mathrm{y}=0$

$x^{2}=9$

$x=\pm 3$

$\text { Thus the points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(3,0) \&(-3,0)$