Question

Find the equation of the parabola with focus $(3,0)$ and directrix $x=-3$.Also find the length of the latus rectum.

Answer 1

The focus of the parabola is $F(3,0)$ and its directrix is the line $x=-3$ i.e., $x+3=0$

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 $FP=MP$

$\Rightarrow \sqrt{{(x-3)}^2+{(y-0)}^2}=\frac{|x+3|}{1}$

$\Rightarrow x^2-6x+9+y^2=x^2+6x+9$

$\Rightarrow -6x+y^2=6x$

$\Rightarrow y^2=12x$ which is the required equation of the parabola.

Comparing it with $y^2=4ax$ we get $4a=12\Rightarrow a=\frac{12}{4}=3$

$\therefore$Length of the latus rectum$=4a=4\times 3=12$