Question

Find the points on the curve $\frac{x^2}{4}+\frac{y^2}{25}=1$ at which the tangents are parallel to the (i) x-axis (ii) y-axis.

Answer 1

(i) The slope of the x-axis is 0.
Now, let (x₁, y₁) be the required point.
$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},\frac{{x_1}^2}{4}+\frac{{y_1}^2}{25}=1...\left(1\right) \\ \mathrm{Now},\frac{x^2}{4}+\frac{y^2}{25}=1 \\ \therefore \frac{2x}{4}+\frac{2y}{25}\frac{dy}{dx}=0 \\ \Rightarrow \frac{2y}{25}\frac{dy}{dx}=\frac{-x}{2} \\ \Rightarrow \frac{dy}{dx}=\frac{-25x}{4y} \\ \mathrm{Now}, \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{-25x_1}{4y_1} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=Slope of the}\mathit{\text{x}}\text{-axis [Given]} \\ \therefore \frac{-25x_1}{4y_1}=0 \\ \Rightarrow x_1=0 \\ \mathrm{Also}, \\ 0+\frac{{y_1}^2}{25}=1[\text{From eq. (1)]} \\ \Rightarrow {y_1}^2=25 \\ \Rightarrow y_1=\pm 5 \\ \text{Thus, the required points are}\left(0,5\right)\text{and}\left(0,-5\right)\text{.} \end{gathered}$$

(ii) The slope of the y-axis is $\infty$.
Now, let (x₁, y₁) be the required point.
$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},\frac{{x_1}^2}{4}+\frac{{y_1}^2}{25}=1...\left(1\right) \\ \mathrm{Now},\frac{x^2}{4}+\frac{y^2}{25}=1 \\ \therefore \frac{2x}{4}+\frac{2y}{25}\frac{dy}{dx}=0 \\ \Rightarrow \frac{2y}{25}\frac{dy}{dx}=\frac{-x}{2} \\ \Rightarrow \frac{dy}{dx}=\frac{-25x}{4y} \\ \mathrm{Now}, \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{-25x_1}{4y_1} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=Slope of the}\mathit{\text{y}}\text{-axis [Given]} \\ \therefore \frac{-25x_1}{4y_1}=\infty \\ \Rightarrow \frac{4y_1}{-25x_1}=0 \\ \Rightarrow y_1=0 \\ \mathrm{Also}, \\ \frac{{x_1}^2}{4}=1[\text{From eq. (1)]} \\ \Rightarrow {x_1}^2=4 \\ \Rightarrow x_1=\pm 2 \\ \text{Thus, the required points are}\left(2,0\right)\text{and}\left(-2,0\right)\text{.} \end{gathered}$$