Question

Find the equation of the tangent and the normal to the following curve at the indicated point.
$y^2=4ax$ at $(x_1,y_1)$.

Answer 1

$y^2=4ax$

Differentiating both sides w.r.t. $x,$

$\Rightarrow$ $2y\frac{dy}{dx}=4a$

$\Rightarrow$ $\frac{dy}{dx}=\frac{2a}{y}$

Slope of tangent at $(x_1,y_1)={\left(\frac{dy}{dx}\right)}_{(x_1,y_2)}=\frac{2a}{y_1}=m$

Equation of tangent is,

$y-y_1=m(x-x_1)$

$\Rightarrow$ $y-y_1=\frac{2a(x-x_1)}{y_1}$

$\Rightarrow$ $y{y}_1-y_1^2=2ax-2a{x}_1$

$\Rightarrow$ $y{y}_1-4a{x}_1=2ax-2a{x}_1$

$\Rightarrow$ $y{y}_1=2ax+2a{x}_1$

$\Rightarrow$ $y{y}_1=2a(x+x_1)$

Equation of normal is,

$y-y_1=\frac{1}{Slopeoftangent}(x-x_1)$

$\Rightarrow$ $y-y_1=\frac{-1}{m}(x-x_1)$

$\Rightarrow$ $y-y_1=\frac{-y_1}{2a}(x-x_1)$

$\therefore$ Equation of tangent is $y{y}_1=2a(x+x_1)$

$\therefore$ Equation of normal is $y-y_1=\frac{-y_1}{2a}(x-x_1)$