Find the equations of the tangents to the parabola $y^2 + 12x = 0$, from the point $\left(3, 8\right)$.
Equation of parabola: $\;$ $y^2 + 12x = 0$ $\;$ i.e. $\;$ $y^2 = -12x$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives
$4a = -12$ $\implies$ $a = -3$
Let the equation of the required tangent be
$y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (2a)$
Substituting the value of $\;$ $a$ $\;$ in equation $(2a)$ gives
$y = mx - \dfrac{3}{m}$ $\;\;\; \cdots \; (2b)$
Given: $\;$ Equation $(2b)$ is drawn from the point $\left(3, 8\right)$,
$\implies$ $8 = 3m - \dfrac{3}{m}$
i.e. $\;$ $3m^2 - 8m - 3 = 0$
i.e. $\;$ $\left(3m + 1\right) \left(m - 3\right) = 0$
i.e. $\;$ $m = \dfrac{-1}{3}$ $\;$ or $\;$ $m = 3$
Substituing the values of $m$ in equation $(2b)$ gives
when $\;$ $m = \dfrac{-1}{3}$
$y = \dfrac{-1}{3} x - \dfrac{3}{-1/3}$
i.e. $\;$ $3y = -x + 27$
i.e. $\;$ $x + 3y = 27$ $\;\;\; \cdots \; (3a)$
when $\;$ $m = 3$
$y = 3x - \dfrac{3}{3}$
i.e. $\;$ $3x - y = 1$ $\;\;\; \cdots \; (3b)$
Equations $(3a)$ and $(3b)$ are the required equations of tangents.