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Question
Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis ?
Solution
Let (x1, y1) be the required point.
\[\text { Since the point lie on the curve } . \]
\[\text { Hence } {x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0 . . . \left( 1 \right)\]
\[\text { Now }, x^2 + y^2 - 2x - 3 = 0 \]
\[ \Rightarrow 2x + 2y \frac{dy}{dx} - 2 = 0\]
\[ \therefore \frac{dy}{dx} = \frac{2 - 2x}{2y} = \frac{1 - x}{y}\]
\[\text { Now,} \]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{1 - x_1}{y_1}\]
\[\text { Slope of the tangent } = 0 (\text { Given )}\]
\[ \therefore \frac{1 - x_1}{y_1} = 0\]
\[ \Rightarrow 1 - x_1 = 0\]
\[ \Rightarrow x_1 = 1\]
\[\text { From (1), we get }\]
\[ {x_1}^2 + {y_1}^2 - 2 x_1 - 3 = 0\]
\[ \Rightarrow 1 + {y_1}^2 - 2 - 3 = 0\]
\[ \Rightarrow {y_1}^2 - 4 = 0\]
\[ \Rightarrow y_1 = \pm 2\]
\[\text { Hence, the points are }\left( 1, 2 \right)\text { and }\left( 1, - 2 \right).\]
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