Question

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