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 $\frac{-2}{3}$.

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},{x_1}^2+{y_1}^2=13...\left(1\right) \\ \mathrm{Now},x^2+y^2=13 \\ \text{On differentiating both sides w.r.t.}\mathit{\text{x}}\text{, we get} \\ 2x+2y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-x}{y} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{-x_1}{y_1} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{= Slope of the given line [Given]} \\ \Rightarrow \frac{-x_1}{y_1}=\frac{-2}{3} \\ \Rightarrow x_1=\frac{2y_1}{3}...\left(2\right) \\ \text{From eq. (1), we get} \\ {\left(\frac{2y_1}{3}\right)}^2+{y_1}^2=13 \\ \Rightarrow \frac{13{y_1}^2}{9}=13 \\ \Rightarrow {y_1}^2=9 \\ \Rightarrow y_1=\pm 3 \\ \Rightarrow y_1=3\text{or}{y}_1=-3 \\ \mathrm{and} \\ {x}_1=2\text{or}{x}_1=-2[\text{From eq. (2)]} \\ \text{Thus, the required points are}\left(2,3\right)\text{and}\left(-2,-3\right)\text{.} \end{gathered}$$