UPSC NDA Math Model Paper 3

Given: Equation of curve is $x^{2}+y^{2}-2x-4y+1=0$ and tangent to the given curve is parallel to $Y$ - axis. Let point of the contact be $(x_{1}, y_{1})$ As we know that, if a tangent of any curve is parallel to $Y-$ axis then $\frac{dx}{dy}=0$ Now by differentiating equation of curve $x^{2}+y^{2}-2x-4y+1=0$ with respect to y we get, $\Rightarrow 2x \cdot \frac{dx}{dy}+2y-2 \cdot \frac{dx}{dy}-4=0$ $\Rightarrow \frac{dx}{dy}=\frac{2-y}{x-1}$ $\Rightarrow \left[\frac{dx}{dy}\right]_{(x_{1}, y_{1})}=\frac{2-y_{1}}{x_{1}-1}$ $\because \left[\frac{dx}{dy}\right]_{(x_{1}, y_{1})}=0$ $\Rightarrow y_{1}=2$ $\because (x_{1}, 2)$ is the point of contact i.e $x=x_{1}, y=2$ will satisfy the equation $x^{2}+y^{2}-2x-4y+1=0$ i.e $x_{1}^{2}+4-2x_{1}-8+1=0$ $\Rightarrow x_{1}^{2}-2x_{1}-3=0$ $\Rightarrow (x_{1}-3)\cdot(x_{1}+1)=0$ $\Rightarrow x_{1}=-1$ or 3 So, the required points are: (- 1, 2) and (3, 2)