Q94 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Mathematics
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\(2 \sqrt{5}\)
- Correct answerC\(\sqrt{5}\)
- D\(4 \sqrt{5}\)
Solution
Given, equation of parabola is
\(\begin{aligned}
20\left(x^2+y^2-6 x-2 y+10\right) & =(4 x-2 y-5)^2 \\
\Rightarrow \quad\left(x^2+y^2-6 x-2 y+10\right) & =\left(\frac{4 x-2 y-5}{\sqrt{20}}\right)^2 \\
\Rightarrow \quad(x-3)^2+(y-1)^2 & =\left(\frac{4 x-2 y-5}{\sqrt{20}}\right)^2 \quad \ldots (i)
\end{aligned}\)
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 \(=4 a\)
\(=2(2 a)=2 \frac{\sqrt{5}}{2}=\sqrt{5}\)
Hence, the correct answer is option: 'C'.