Question

The eccentricity of the conic $4(2y-x-3{)}^2-9(2x+y-1{)}^2=80$ is :

• A. $2$
• B. $\frac{\sqrt{13}}{3}$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{13}}{2}$

Answer 1

Answer: (B)

$\frac{\sqrt{13}}{3}$

The given equation can also be written as $4{\left(\frac{2y-x-3}{\sqrt{5}}\right)}^2-9{\left(\frac{2x+y-1}{\sqrt{5}}\right)}^2=16$
Let $X=\frac{2y-x-3}{\sqrt{5}}$ and $Y=\frac{2x+y-1}{\sqrt{5}}$
$4X^2-9Y^2=16$
$\Rightarrow \frac{X^2}{4}-\frac{Y^2}{\frac{16}{9}}=1$
Now $b^2=a^2(e^2-1)$
So, $\frac{16}{9}=4(e^2-1)$
$\Rightarrow e=\frac{\sqrt{13}}{3}$
Hence, option 'B' is correct.