Question

Find the equation of the parabola with focus $(6,0)$ and directrix $x=-6$.Also find the length of the latus rectum.

Answer 1

The focus of the parabola is $F(6,0)$ and its directrix is the line $x=-6$ i.e., $x+6=0$

Let $P(x,y)$ be any point in the plane of directrix and focus, and $MP$ be the perpendicular distance from $P$ to the directrix,then $P$ lies on parabola iff $FP=MP$

$\Rightarrow \sqrt{{(x-6)}^2+{(y-0)}^2}=\frac{|x+6|}{1}$

$\Rightarrow x^2-12x+36+y^2=x^2+12x+36$

$\Rightarrow -12x+y^2=12x$

$\Rightarrow y^2=24x$ which is the required equation of the parabola.

Comparing it with $y^2=4ax$ we get $4a=24\Rightarrow a=\frac{24}{4}=6$

$\therefore$Length of the latus rectum$=4a=4\times 6=24$