#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 9 Maths Textbook Solution.

Answer: $\text { The required point is }(1,0) \&(1,4)$

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

Given:$\text { The curve is } x^{2}+y^{2}-2 x-4 y+1=0$

Solution: $x^{2}+y^{2}-2 x-4 y+1=0$

$\text { Differentiating the above with respect to } x$

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

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

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

$\frac{d y}{d x}(2 y-4)=-2 x+2$

$\frac{d y}{d x}=\frac{-(x-1)}{(y-2)} \rightarrow(1)$

$\frac{d y}{d x}=\text { The slope of tangent }=\tan \theta$

$\text { Since the tangent is parallel to the } \mathrm{x} \text { -axis }$

$\frac{d y}{d x}=\tan (0)=0 \quad \rightarrow(2)$

From (1)&(2) we get

$\frac{-(x-1)}{(y-2)}=0$

$-(x-1)=0$

$-x+1=0$

$x=1$

$\text { Substituting } x=1 \text { in } x^{2}+y^{2}-2 x-4 y+1=0, \text { we get }$

$\text { (1) }^{2}+y^{2}-2(1)-4 y+1=0$

$1+y^{2}-2-4 y+1=0$

$y^{2}-4 y=0$

$y(y-4)=0$

$y=0 \& y-4=0$

$y=0 \& y=4$

$\text { So, the required point is }(1,0) \&(1,4)$