Question

Find the equations of the tangents to the parabola $y^2=16x$, which are parallel and perpendicular respectively to the line 2x-y+5=0. Find also the coordinates of their points of contact.

Answer 1

We have,

Equation of parabola is $y^2=16x......\left(1\right)$

And equation of tangent is

$2x-y+5=0$

$y=2x+5.......\left(2\right)$

Tangent to the parallel and perpendicular to the $2x-y+5=0$

From equation (1)

$y^2=16x$

On comparing that

$y^2=4ax$

Now, $a=4$

We know that,

Any tangent to $y^2=4ax$ is $y=mx+\frac{a}{m}$

Here,

$y=mx+\frac{4}{m}$

From equation (2)

$y=2x+5$

On comparing that,

$y=mx+c$

For Parallel $slope\left(m\right)=2$

For perpendicular $Slope\left(m\right)=-\frac{1}{m}=-\frac{1}{2}$

Then, required equation of tangent

For slope $m=2$ is

$y=mx+\frac{4}{m}$

$y=2x+\frac{4}{2}$

$y=2x+2$

$2x-y-2=0$

For slope $m=-\frac{1}{2}$

$y=mx+\frac{4}{m}$

$y=-\frac{1}{2}x+\frac{4}{-\frac{1}{2}}$

$y=-\frac{1}{2}x-8$

$2y=-x-16$

$x+2y+16=0......\left(3\right)$

Now

Point of contact is

From equation (2) and (3) to, and we get,

$y=2x+5......\left(2\right)$

$x+2y+16=0......\left(3\right)$

$So,$

$x+2(2x+5)+16=0$

$\Rightarrow x+4x+10+16=0$

$\Rightarrow 5x+26=0$

$\Rightarrow x=-\frac{26}{5}$

Put the value of x in equation (2) and we get,

$y=2x+5$

$y=2(-\frac{26}{5})+5$

$y=\frac{-52+25}{5}$

$y=-\frac{27}{5}$

Hence, the point of contact is $(\frac{-26}{5},\frac{-27}{5})$.

Hence, this is the answer.