Question

Find the points on the curve x² + y² = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7 ?

Answer 1

Let (x₁, y₁) represent the required point.
The slope of line 2x + 3y = 7 is \[\frac{- 2}{3}\] .

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

\[\text { Hence }, {x_1}^2 + {y_1}^2 = 13 . . . \left( 1 \right)\]

\[\text { Now }, x^2 + y^2 = 13\]

\[\text { On differentiating both sides w.r.t.x, we get}\]

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

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

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

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \text { Slope of the given line [Given] }\]

\[ \Rightarrow \frac{- x_1}{y_1} = \frac{- 2}{3}\]

\[ \Rightarrow x_1 = \frac{2 y_1}{3} . . . \left( 2 \right)\]

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

\[ \left( \frac{2 y_1}{3} \right)^2 + {y_1}^2 = 13\]

\[ \Rightarrow \frac{13 {y_1}^2}{9} = 13\]

\[ \Rightarrow {y_1}^2 = 9\]

\[ \Rightarrow y_1 = \pm 3\]

\[ \Rightarrow y_1 = 3 or y_1 = - 3\]

\[\text { and }\]

\[ x_1 = 2 or x_1 = - 2 [\text { From eq.} (2)]\]

\[\text {Thus, the required points are }\left( 2, 3 \right)\text { and }\left( - 2, - 3 \right).\]