Find the equation of the two tangents that can be drawn from the point $\left(2,-3\right)$ to the parabola $y^2 = 4x$.
Equation of parabola is $\;$ $y^2 = 4x$ $\;\;\; \cdots \; (1)$
Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives $\;$ $a = 1$
Let the equation of tangent be $\;$ $y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (2)$
Substituting the value of $a$ in equation $(2)$ gives
$y = mx + \dfrac{1}{m}$ $\;\;\; \cdots \; (3)$
Equation $(3)$ passes through the point $\left(2, -3\right)$.
$\therefore$ $\;$ We have $\;$ $-3 = 2m + \dfrac{1}{m}$
i.e. $\;$ $2m^2 + 3m + 1 = 0$
i.e. $\;$ $\left(2m + 1\right) \left(m + 1\right) = 0$
i.e. $\;$ $m = -1$ $\;$ or $\;$ $m = \dfrac{-1}{2}$
When $m = -1$, we have from equation $(3)$, $\;$ $y = -x -1$
i.e. $\;$ $x + y + 1 = 0$ $\;\;\; \cdots \; (4a)$
When $m = \dfrac{-1}{2}$, we have from equation $(3)$, $\;$ $y = \dfrac{-x}{2} - \dfrac{1}{1 / 2}$
i.e. $\;$ $y = \dfrac{-x}{2} - 2$
i.e. $\;$ $x + 2y + 4 = 0$ $\;\;\; \cdots \; (4b)$
Equations $(4a)$ and $(4b)$ are the required equations of tangents.