Find the equations of tangent and normal to the hyperbola $9x^2 - 5y^2 = 31$ at $\left(2,-1\right)$.
Equation of given hyperbola: $\;$ $9x^2 - 5y^2 = 31$
i.e. $\;$ $\dfrac{x^2}{31 / 9} - \dfrac{y^2}{31 / 5} = 1$ $\;\;\; \cdots \; (1)$
The transverse axis of the hyperbola is along the X axis.
Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives
$a^2 = \dfrac{31}{9}$ $\;$ and $\;$ $b^2 = \dfrac{31}{5}$
Equation of tangent to the hyperbola at the point $\left(x_1, y_1\right)$ is
$\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$
Given: $\left(x_1, y_1\right) = \left(2, -1\right)$
$\therefore$ $\;$ Required equation of tangent is
$\dfrac{2x}{31 / 9} + \dfrac{y}{31 / 5} = 1$
i.e. $\;$ $18x + 5y = 31$
Equation of normal to the hyperbola at the point $\left(x_1, y_1\right)$ is
$\dfrac{a^2 x}{x_1} + \dfrac{b^2 y}{y_1} = a^2 + b^2$
$\therefore$ $\;$ Required equation of normal is
$\dfrac{31 x}{9 \times 2} + \dfrac{31 y}{5 \times \left(-1\right)} = \dfrac{31}{9} + \dfrac{31}{5}$
i.e. $\;$ $\dfrac{x}{18} - \dfrac{y}{5} = \dfrac{14}{45}$
i.e. $\;$ $5x - 18 y - 28 = 0$