Question

The length of the latusrectum of the parabola
$20\left(x^2+y^2-6x-2y+10\right)=(4x-2y-5{)}^2$, is

• A. $\frac{\sqrt{5}}{2}$
• B. $2\sqrt{5}$
• C. $\sqrt{5}$
• D. $4\sqrt{5}$

Answer 1

Answer: (C)

Given, equation of parabola is
$20\left(x^2+y^2-6x-2y+10\right)=(4x-2y-5{)}^2$
$\Rightarrow \left(x^2+y^2-6x-2y+10\right)={\left(\frac{4x-2y-5}{\sqrt{20}}\right)}^2$
$\Rightarrow (x-3{)}^2+(y-1{)}^2={\left(\frac{4x-2y-5}{\sqrt{20}}\right)}^2$
In Eq. (i), focus is $(3,1)$ and equation of directrix is $4x-2y-5=0$
So, distance from focus to directrix is $\frac{|12-2-5|}{\sqrt{20}}$
$=\frac{\sqrt{5}}{2}=2a$
Now, length of latursrectum $=4a$
$=2(2a)=2\frac{\sqrt{5}}{2}=\sqrt{5}$