#### Question

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

#### Solution

Let (x_{1}, y_{1}) be the required point.

\[\text { Since the point lie on the curve } . \]

\[\text { Hence } {x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0 . . . \left( 1 \right)\]

\[\text { Now }, x^2 + y^2 - 2x - 3 = 0 \]

\[ \Rightarrow 2x + 2y \frac{dy}{dx} - 2 = 0\]

\[ \therefore \frac{dy}{dx} = \frac{2 - 2x}{2y} = \frac{1 - x}{y}\]

\[\text { Now,} \]

\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{1 - x_1}{y_1}\]

\[\text { Slope of the tangent } = 0 (\text { Given )}\]

\[ \therefore \frac{1 - x_1}{y_1} = 0\]

\[ \Rightarrow 1 - x_1 = 0\]

\[ \Rightarrow x_1 = 1\]

\[\text { From (1), we get }\]

\[ {x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0\]

\[ \Rightarrow 1 + {y_1}^2 - 2 - 3 = 0\]

\[ \Rightarrow {y_1}^2 - 4 = 0\]

\[ \Rightarrow y_1 = \pm 2\]

\[\text { Hence, the points are }\left( 1, 2 \right)\text { and }\left( 1, - 2 \right).\]