Question

Find the points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Answer 1

(i) The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},\frac{{x_1}^2}{9}+\frac{{y_1}^2}{16}=1...\left(1\right) \\ \mathrm{Now}, \\ \frac{x^2}{9}+\frac{y^2}{16}=1 \\ \Rightarrow \frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0 \\ \Rightarrow \frac{y}{16}\frac{dy}{dx}=\frac{-x}{9} \\ \Rightarrow \frac{dy}{dx}=\frac{-16x}{9y} \\ \mathrm{Now}, \\ \text{Slope of the tangent at}\left(x,y\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{-16x_1}{9y_1} \\ \text{Slope of the tangent at}\left(x,y\right)\text{= Slope of the}\mathit{\text{x}}\text{-axis [Given]} \\ \therefore \frac{-16x_1}{9y_1}=0 \\ \Rightarrow x_1=0 \\ 0+\frac{{y_1}^2}{16}=1[\text{From eq. (1)]} \\ \mathrm{Also}, \\ {y_1}^2=16 \\ {y}_1=\pm 4 \\ \text{Thus, the required points are}\left(0,4\right)\text{and}\left(0,-4\right)\text{.} \end{gathered}$$

(ii) The slope of the y-axis is $\infty$.
Let (x₁, y₁) be the required point.
Given:
$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},\frac{{x_1}^2}{9}+\frac{{y_1}^2}{16}=1...\left(1\right) \\ \frac{x^2}{9}+\frac{y^2}{16}=1 \\ \Rightarrow \frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0 \\ \Rightarrow \frac{y}{16}\frac{dy}{dx}=\frac{-x}{9} \\ \Rightarrow \frac{dy}{dx}=\frac{-16x}{9y} \\ \mathrm{Now}, \\ \text{Slope of the tangent at}\left(x,y\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{-16x_1}{9y_1} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{= Slope of the}\mathit{\text{y}}\text{-axis [Given]} \\ \therefore \frac{-16x_1}{9y_1}=\infty \\ \Rightarrow \frac{9y_1}{-16x_1}=0 \\ \Rightarrow y_1=0 \\ \Rightarrow \frac{{x_1}^2}{9}+0=1[\text{From eq. (1)]} \\ \Rightarrow {x_1}^2=9 \\ \Rightarrow x_1=\pm 3 \\ \text{Thus, the required points are}\left(3,0\right)\text{and}\left(-3,0\right)\text{.} \end{gathered}$$