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

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Solution

Given equation of curve which is
$x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \Rightarrow 2 x + 2 y \frac{d t}{d x} - 2 - 4 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} (2 y - 4) = 2 - 2 x \Rightarrow \frac{d y}{d x} = \frac{2 (1 - x)}{2 (y - 2)}$
Since, the tangents are parallel to the axis i.e., $t a n θ = t a n 90^{\circ} = \frac{d y}{d x} ∴ \frac{1 - x}{y - 2} = \frac{1}{0} \Rightarrow y - 2 = 0 \Rightarrow y = 2$
For y = 2 from Eq. (i), we get
$x^{2} + 2^{2} - 2 x - 4 \times 2 + 1 = 0 \Rightarrow x^{2} - 2 x - 3 = 0 \Rightarrow x^{2} - 3 x + x - 3 = 0 \Rightarrow x (x - 3) + 1 (x - 3) = 0 \Rightarrow (x + 1) (x - 3) = 0 ∴ x = - 1, x = 3$
So, the required points are (-1, 2) and (3,2).

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