Question

At what points on the curve ${\text{x}}^2+{\text{y}}^2-2\text{x}-4\text{y}+1=0$, are the tangents parallel to the y-axis

Answer 1

Given curve:
${\text{x}}^2+{\text{y}}^2-2\text{x}-4\text{y}+1=0$...(1)
Differentiating both sides w.r.t x, we get
$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-2-4\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}(\text{y}-2)=1-\text{x}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{y}-\text{z}}$...(2)
Slope,$\frac{\text{dy}}{\text{dx}}=\text{tan}\frac{\pi }{2}=\frac{1}{0}$
$\Rightarrow \frac{1-\text{x}}{\text{y}-\text{z}}=\frac{1}{0}$ [From eq. (2)]
$\Rightarrow \text{y}-2=0$
$\Rightarrow \text{y}=2$...(3)
$\Rightarrow {\text{x}}^2+4-2\text{x}-8+1=0$
$\Rightarrow {\text{x}}^2-2\text{x}-3=0$
$\Rightarrow (\text{x}-3)(\text{x}+1)=0$
Hence, the points are $(3,2)$ and $(-1,2)$
Hence, at $(3,2)$ and $(-1,2)$ on the given curve, the tangents are parallel to the y-axis.