Q98 / 160 · AP EAMCET 2016 · Mathematics
The equations of the latusrectum of the ellipse $9 x^2+25 y^2-36 x+50 y-164=0$ are
- A$x-4=0, x+2=0$
- Correct answerB$x-6=0, x+2=0$
- C$x+6=0, x-2=0$
- D$x+4=0, x+5=0$
Solution
We have,
$\begin{aligned}
& \Rightarrow 9 x^2+25 y^2-36 x+50 y-164=0 \\
& \Rightarrow 9 x^2-36 x+36++25 y^2+50 y-25=164+36+25 \\
& \Rightarrow 9\left(x^2-4 x+4\right)-25\left(y^2+2 y-1\right)=225 \\
& \Rightarrow 9(x-2)^2-25(y+1)^2=225
\end{aligned}$
$\begin{aligned}
& \Rightarrow \frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1 \\
& \text { Eccentricity e }=\sqrt{1-\left(\frac{b}{a}\right)^2} \\
& =\sqrt{1-\frac{9}{25}}=\frac{4}{5}
\end{aligned}$
Equation of the latusrectum,
$\begin{aligned}
& x-2= \pm a e \\
& \Rightarrow \quad x-2= \pm 5 \times \frac{4}{5}= \pm 4 \\
& \Rightarrow \quad x= \pm 4+2 \\
& \Rightarrow \quad x=6 \text { and } x=-2 \\
& \therefore \quad x-6=0, x+2=0
\end{aligned}$
Hence, the correct answer is option: 'B'.