Find the equation of the tangent and normal at $\left(-2, \dfrac{1}{4}\right)$ to the rectangular hyperbola $2xy - 2x - 8y - 1 = 0$
Equation of given rectangular hyperbola is
$2xy - 2x - 8y -1 = 0$
i.e. $\;$ $2xy - 2x - 8y = 1$
i.e. $\;$ $2xy - 2x - 8y + 8 = 9$
i.e. $\;$ $xy - x - 4y + 4 = \dfrac{9}{2}$
i.e. $\;$ $x \left(y - 1\right) - 4 \left(y - 1\right) = \dfrac{9}{2}$
i.e. $\;$ $\left(x - 4\right) \left(y - 1\right) = \dfrac{9}{2}$ $\;\;\; \cdots \; (1)$
The standard equation of rectangular hyperbola with center at $\left(h, k\right)$ is:
$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (2)$
Comparing equations $(1)$ and $(2)$ we have
Center $= \left(h, k\right) = \left(4, 1\right)$ $\;$ and $\;$ $c^2 = \dfrac{9}{2}$
Equation of tangent to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is
$x \left(y_1 - k\right) + y \left(x_1 - h\right) = c^2 - hk + x_1y_1$
Given: Point $\left(x_1, y_1\right) = \left(-2, \dfrac{1}{4}\right)$
$\therefore$ $\;$ The required equation of tangent is
$x \left(\dfrac{1}{4} - 1\right) + y \left(-2 - 4\right) = \dfrac{9}{2} - \left(4 \times 1\right) + \left(-2 \times \dfrac{1}{4}\right)$
i.e. $\;$ $\dfrac{-3}{4}x - 6y = 0$
i.e. $\;$ $\dfrac{-x}{4}- 2y = 0$
i.e. $\;$ $x + 8y = 0$
Equation of normal to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is
$x \left(x_1 - h\right) - y \left(y_1 - k\right) = x_1 \left(x_1 - h\right) - y_1 \left(y_1 - k\right)$
$\therefore$ $\;$ The required equation of normal is
$x \left(-2 - 4\right) - y \left(\dfrac{1}{4} - 1\right) = -2 \left(-2 - 4\right) - \dfrac{1}{4} \left(\dfrac{1}{4} - 1\right)$
i.e. $\;$ $- 6x + \dfrac{3}{4}y = \dfrac{195}{16}$
i.e. $\;$ $- 2x + \dfrac{y}{4} = \dfrac{65}{16}$
i.e. $\;$ $32x - 4y + 65 = 0$