Find the equation of the normal to the parabola $y^2 = 4x$ which is perpendicular to the line $2x + 6y + 5 = 0$.
Equation of parabola: $\;$ $y^2 = 4x$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives
$4a = 4$ $\implies$ $a = 1$
Equation of line: $\;$ $2x + 6y + 5 = 0$ $\;\;\; \cdots \; (2)$
Slope of equation $(2)$ is $\;$ $m_1 = \dfrac{-2}{6} = \dfrac{-1}{3}$
The required normal is perpendicular to line $(2)$.
$\therefore \;$ Slope of normal $= m = \dfrac{-1}{m_1} = 3$
Let the equation of the required normal be $\;$ $y = mx - 2am - am^2$ $\;\;\; \cdots \; (3)$
Substituting the values of $a$ and $m$ in equation $(3)$ gives
$y = 3x - 2 \times 1 \times 3 - 1 \times 3^3$
i.e. $\;$ $3x - y - 33 = 0$ $\;\;\; \cdots \; (4)$
Equation $(4)$ is the equation of the required normal.