Find the equations of tangent and normal to the parabola $y^2 = 8x$ at $t = \dfrac{1}{2}$.
Given: Equation of parabola is $\;$ $y^2 = 8x$
Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives
$4a = 8 \implies a = 2$
Equation of tangent to a parabola at a point $t$ is $\;$ $yt = x+ at^2$
Equation of normal to a parabola at a point $t$ is $\;$ $y + tx = 2at + at^3$
Given: $\;$ $t = \dfrac{1}{2}$
$\therefore$ $\;$ Required equation of tangent is:
$y \times \dfrac{1}{2} = x + 2 \times \dfrac{1}{4}$
i.e. $\;$ $2x - y + 1 = 0$
Required equation of normal is:
$y + \dfrac{x}{2} = 2 \times 2 \times \dfrac{1}{2} + 2 \times \dfrac{1}{8}$
i.e. $\;$ $y + \dfrac{x}{2} = \dfrac{9}{4}$
i.e. $\;$ $2x + 4y - 9 = 0$