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Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 18 Sub Question 1 Maths Textbook Solution.

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Answer: $\text { The points at which the tangents are parallel to } \mathrm{x} \text { -axis are }(1,2) \&(1,-2)$

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

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

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

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

$\begin{aligned} &2 y \frac{d y}{d x}=2-2 x \\ &2 y \frac{d y}{d x}=2(1-x) \\ &\frac{d y}{d x}=\frac{1-x}{y} \end{aligned}$

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

$\text { substitute } x=1 \operatorname{in} x^{2}+y^{2}-2 x-3=0$

$(1)^{2}+y^{2}-2(1)-3=0$

$1+y^{2}-2-3=0$

$1+y^{2}-5=0$

$y^{2}=5-1$

$y^{2}=4$

$y=\pm 2$

$\text { Thus the points at which the tangents are parallel to } \mathrm{x} \text { -axis are }(1,2) \&(1,-2)$

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