Q92 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics
The length of the latusrectum of the parabola $(x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2$ is
- A$\frac{1}{5}$
- Correct answerB$\frac{2}{5}$
- C$\frac{3}{5}$
- D$\frac{4}{5}$
Solution
Given, equation of parabola is
$$
\begin{aligned}
& (x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2 \\
& \Rightarrow 25\left[(x-2)^2+(y-3)^2\right]=(3 x-4 y+7)^2
\end{aligned}
$$
Focus of the parabola $=(2,3)$
Equation of directrix $=3 x-4 y+7$
Length of latus rectum is twice the perpendicular distance from focus to directrix.
$$
L=\frac{2|3 \times 2-4 \times 3+7|}{\sqrt{3^2+(-4)^2}}=2\left[\frac{6-12+7}{5}\right]=\frac{2}{5}
$$
Hence, the correct answer is option: 'B'.