#### Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 18 Sub Question 2 Maths Textbook Solution.

Answer: $\text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(-1,0) \&(3,0)$

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

Given: $x^{2}+y^{2}-2 x-3=0 \quad \rightarrow(1)$

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 \\ &2 x^{2-1}+2 y^{2-1} \frac{d y}{d x}-2=0 \\ &2 x+2 y \frac{d y}{d x}=2 \\ &2 y \frac{d y}{d x}=2-2 x \\ &2 y \frac{d y}{d x}=2(1-x) \\ &\frac{d y}{d x}=\frac{1-x}{y} \end{aligned}

\begin{aligned} &\text { Now the tangent is parallel to the } \mathrm{y} \text { -axis if the slope of the normal is zero }\\ &\text { Slope of normal }=\frac{-1}{\text { Slope of tangent }} \Rightarrow \frac{-1}{\frac{1-x}{y}}=\frac{-y}{1-x}=0 \end{aligned}

But

\begin{aligned} &x^{2}+y^{2}-2 x-3=0 \text { for } y=0 \\ &x^{2}+0-2 x-3=0 \\ &x^{2}-2 x-3=0 \\ &x^{2}-3 x+x-3=0 \\ &x(x-3)+1(x-3)=0 \\ &x+1=0 \quad \& x-3=0 \\ &x=-1 \quad \& x=3 \end{aligned}

Thus $\text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(-1,0) \&(3,0)$