Find the points on the curve $x^{2} + y^{2} - 2 x - 3 = 0$ at which tangents are parallel to the X-axis.

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Solution

The equation of the given curve is $x^{2} + y^{2} - 2 x - 3 = 0$

On differentiating w.r.t. x, we get

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

For tangent parallel to X-axis, we must have, $\frac{d y}{d x} = 0$

$\Rightarrow \frac{1 - x}{y} = 0 \Rightarrow 1 - x = 0 \Rightarrow x = 1$

Substituting x=1 in Eq. (i), we get

$1^{2} + y^{2} - 2 \times 1 - 3 = 0 \Rightarrow y^{2} - 4 = 0 \Rightarrow y = \pm 2$

Hence, the point at which the tangents are parallel to the X-axis are (1,2) and (1,-2).

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