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

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Solution

Given curve is, $x^{2} + y^{2} 2 x - 3 = 0$ ...(1)
Differentiating w.r.t. $x \frac{d (x^{2} + y^{2} - 2 x - 3)}{d x} 0$
$\Rightarrow 2 x + 2 y \times \frac{d y}{d x} - 2 = 0$
$\Rightarrow 2 y \times \frac{d y}{d x} = 2 - 2 x$
$\Rightarrow \frac{d y}{d x} = \frac{2 - 2 x}{2 y}$
$\Rightarrow \frac{d y}{d x} = \frac{1 - x}{y}$ ...(2)
Tangents are parallel to the x-axis, then the slope $= 0$
$\frac{d y}{d x} = 0$
$\Rightarrow \frac{1 - x}{y} = 0$
$\Rightarrow x = 1$
Putting $x = 1$ in equation (1)
$(1)^{2} + y^{2} - 2 (1) - 3 = 0$
$\Rightarrow 1 + y^{2} - 2 - 3 = 0$
$\Rightarrow y^{2} - 4 = 0$
$\Rightarrow y^{2} = 4$
$\Rightarrow y = \pm 2$
Hence, the required points are $(1, 2) & (1, - 2)$ .

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