Question

The length of the latus rectum of the parabola $25\left[\right(x-2{)}^2+(y-4{)}^2]=(4x-3y+12{)}^2$ is

• A. $\frac{16}{5}$
• B. $\frac{12}{5}$
• C. $\frac{8}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A $\frac{16}{5}$

The given equation of the parabola can be written as
$(x-2{)}^2+(y-4{)}^2={\left(\frac{4x-3y+12}{\sqrt{(4{)}^2+(-3{)}^2}}\right)}^2$
$\therefore$ The coordinates of focus are $(2,4)$ and the equation of directrix is $4x-3y+12=0$.
The distance of the focus from the directrix
$=\left|\frac{4\left(2\right)-3\left(4\right)+12}{\sqrt{4^2+(-3{)}^2}}\right|=\frac{8}{5}$
$\therefore$ The length of latus rectum
$=2\times$ distance of the focus from the directrix
$=2\times \frac{8}{5}=\frac{16}{5}$