Question

TANGENT AND NORMAL
(i) Find the points on the hyperbola $2x^2-3y^2=6$ at which the slope of the tangent line is (-1).

Answer 1

$$\begin{gathered} 2{\mathrm{x}}^2-3{\mathrm{y}}^2=6 \\ \mathrm{For}\mathrm{slope},\mathrm{differentiating}\mathrm{wrt}\mathrm{x} \\ 4\mathrm{x}-6\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{x}}{3\mathrm{y}} \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{slope}=\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=-1 \\ \frac{2\mathrm{x}}{3\mathrm{y}}=-1 \\ 2\mathrm{x}=-3\mathrm{y} \\ \mathrm{so}\mathrm{putting}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}\mathrm{of}\mathrm{hyperbola} \\ 2\times {\left(\frac{-3\mathrm{y}}{2}\right)}^2-3{\mathrm{y}}^2=6 \\ 9{\mathrm{y}}^2-6{\mathrm{y}}^2=12 \\ {\mathrm{y}}^2=4,\mathrm{y}=\pm 2 \\ \mathrm{so}2{\mathrm{x}}^2-3\times 4=6 \\ 2{\mathrm{x}}^2=18 \\ \mathrm{x}=\pm 3 \\ \mathrm{so}\mathrm{the}\mathrm{points}\mathrm{are}\left(\pm 3,\pm 2\right) \end{gathered}$$