#### At what points on the curve $x^2 + y^2 - 2x - 4y + 1 = 0$, the tangents are parallel to the y-axis?

Given: equation of a curve $x^2 + y^2 - 2x - 4y + 1 = 0$

To find: the points on the curve $x^2 + y^2 - 2x - 4y + 1 = 0$, the tangents are parallel to the y-axis

Explanation: the given equation of curve as $x^2 + y^2 - 2x - 4y + 1 = 0$

Differentiating the equation with respect to X

\begin{aligned} &\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}+1\right)}{\mathrm{dx}}=\frac{\mathrm{d}(0)}{\mathrm{dx}}\\ &\text { Now applying the sum rule of differentiation }\\ &\Rightarrow \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}-\frac{\mathrm{d}(4 \mathrm{y})}{\mathrm{dx}}+\frac{\mathrm{d}(1)}{\mathrm{dx}}=0\\ &\Rightarrow 2 x+2 y \frac{d(y)}{d x}-2-4 \frac{d y}{d x}+0=0\\ &\Rightarrow(2 y-4) \frac{d(y)}{d x}=2-2 x\\ &\Rightarrow \frac{d y}{d x}=\frac{2(1-x)}{2(y-2)} \end{aligned}

Sense the tangents are parallel to the axis as mentioned in the question

Thus,

\begin{aligned} &\tan \theta=\tan 90^{\circ}=\frac{\mathrm{dy}}{\mathrm{dx}}\\ &\text { Value from equation (i) can be substituted equation }\\ &\frac{1}{0}=\frac{(1-x)}{(y-2)} \end{aligned}

$\\ { \Rightarrow y-2 = 0}\\ { \Rightarrow y = 2}\\$

Putting y = 2 in curve equation, we get

$\\x\textsuperscript2 + y^2 - 2x - 4y + 1 = 0\\$

$\\ \Rightarrow x\textsuperscript2 + 2^2 - 2x - 4(2) + 1 = 0\\$

$\\\Rightarrow x\textsuperscript2 + 4 - 2x - 8 + 1 = 0\\ \Rightarrow x\textsuperscript2- 2x-3 = 0\\$

After the splitting of the middle term,

$\\ \Rightarrow x\textsuperscript{2}-3x+x-3 = 0\\ { \Rightarrow x(x-3)+1(x-3) = 0}\\ { \Rightarrow (x+1)(x-3) = 0}\\ { \Rightarrow x+1 = 0 or x-3 = 0}\\ { \Rightarrow x = -1 or x = 3}\\$

Thus, the needed points are(-1, 2) and (3, 2).

Hence the points on the curve $x^2 + y^2 - 2x - 4y + 1 = 0$, the tangents are parallel to the y-axis are (-1, 2) and (3, 2).