Question

Find the points on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x-axis.

Answer 1

Let the required point be (x₁, y₁).
Slope of the tangent at this point = tan 45$°$ = 1
Given:
$$\begin{gathered} xy+4=0...\left(1\right) \\ \mathrm{Since}\mathrm{the}\mathrm{point}\mathrm{satisfies}\mathrm{the}\mathrm{above}\mathrm{equation}, \\ {x}_1{y}_1+4=0...\left(2\right) \\ \text{On differentiating equation}\left(2\right)\text{both sides with respect to}x,\mathrm{we}\mathrm{get} \\ x\frac{dy}{dx}+y=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-y}{x} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x,y\right)}=\frac{-y_1}{x_1} \\ \text{Slope of the tangent =1 [Given]} \\ \therefore \frac{-y_1}{x_1}=1 \\ \Rightarrow x_1=-y_1 \\ \text{On substituting the value of}{x}_1\text{in eq. (2), we get} \\ -{y_1}^2+4=0 \\ \Rightarrow {y_1}^2=4 \\ \Rightarrow y_1=\pm 2 \\ \text{Case 1} \\ \mathrm{When}{y}_1=2,x_1=-y_1=-2 \\ \text{∴ (}{x}_1,y_1\text{) = (-2, 2)} \\ \text{Case 2} \\ \mathrm{When}{y}_1=-2,x_1=-y_1=2 \\ \text{∴}\left(x_1,y_1\right)\text{= (2, -2)} \end{gathered}$$