Question

Find the equation of the normal to the parabola $y^2=4x$, which is
(i) parallel to the line $y=2x-5$,
(ii) perpendicular to the line $x+3y+1=0$.

Answer 1

Slope of tangent to parabola is $2y\frac{dy}{dx}=4$

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

$\therefore$ Slope of normal $m=-\frac{1}{\frac{dy}{dx}}=-\frac{y}{2}$

$\left(i\right)$ parallel to the line $y=2x-5$

$\therefore m=2$ ... ($\because$ Slope will be same)

$m=-\frac{y}{2}=2\to y=-4$

Putting value of $y$ in equation of parabola we get $x=4$.

So point on parabola for the normal is $(4,-4)$

$\therefore$ equation is $y-(-4)=2(x-4)$

$2x-y=12$

$\left(ii\right)$ perpendicular to the line $x+3y+1=0$

$\therefore m=3$ ....... $(\because$ Slope will be= -$\frac{1}{Slopeofperpendicularline}$)

$m=-\frac{y}{2}=3\to y=-6$

Putting value of $y$ in equation of parabola we get $x=9$.

So point on parabola for the normal is $(9,-6)$

$\therefore$ equation is $y-(-6)=3(x-9)$

$3x-y=33$