Find the points on the curve


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


Let $\left(x_{1}, y_{1}\right)$ be the required point.

Since the point lie on the curve.

Hence $x_{1}{ }^{2}+y_{1}{ }^{2}-2 x_{1}-3=0 \ldots$ (1)

Now, $x^{2}+y^{2}-2 x-3=0$

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

$\therefore \frac{d y}{d x}=\frac{2-2 x}{2 y}=\frac{1-x}{y}$


Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1-x_{1}}{y_{1}}$

Slope of the tangent $=0 \quad$ (Given)

$\therefore \frac{1-x_{1}}{y_{1}}=0$

$\Rightarrow 1-x_{1}=0$

$\Rightarrow x_{1}=1$

From (1), we get

$x_{1}^{2}+y_{1}^{2}-2 x_{1}-3=0$

$\Rightarrow 1+y_{1}^{2}-2-3=0$

$\Rightarrow y_{1}^{2}-4=0$

$\Rightarrow y_{1}=\pm 2$

Hence, the points are $(1,2)$ and $(1,-2)$.

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