Question

At what point the slope of the tangent to the curve x² + y² − 2x − 3 = 0 is zero

विकल्प

• (3, 0), (−1, 0)

• (3, 0), (1, 2)

• (−1, 0), (1, 2)

• (1, 2), (1, −2)

Answer 1

(1, 2), (1, −2)

 

Let (x₁, y₁) be the required point.

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

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

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

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

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

\[\text { Now}, \]

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

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

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

\[ \Rightarrow 1 - x_1 = 0\]

\[ \Rightarrow x_1 = 1\]

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

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

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

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

\[ \Rightarrow y_1 = \pm 2\]

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