#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 17 Sub Question 1 Maths Textbook Solution.

Answer: $\text { The points at which the tangents are parallel to } \mathrm{x} \text { -axis are }(0,5) \&(0,-5)$

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

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

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

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

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

$\frac{2 y}{25} \frac{d y}{d x}=\frac{-x}{2}$

$\frac{d y}{d x}=\frac{-x}{2} \times \frac{25}{2 y}$

$\frac{d y}{d x}=\frac{-25 x}{4 y}$

Now the tangent is parallel to the x-axis if the slope of the tangent is zero

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

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

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

\begin{aligned} &\frac{0}{4}+\frac{y^{2}}{25}=1 \\ &y^{2}=25 \\ &y=\pm 5 \end{aligned}

$\text { Substitute } y=\pm 5 \text { in } \frac{-25 x}{4 y}$

$\begin{array}{ll} \text { Put } y=5 & , \quad \text { Put } y=-5 \\ \frac{-25 x}{4(5)}=0 & , \quad \frac{-25 x}{4(-5)}=0 \\ \frac{-5 x}{4}=0 & , \quad \frac{5 x}{4}=0 \end{array}$

$x=0 \quad, \quad x=0$

Thus the points at which the tangents are parallel to x-axis are 0,5&0,-5