Question

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

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

Answer 1

The correct option is D $\frac{\sqrt{13}}{3}$

Here, $2y-x-3$ and $2x+y-1$ are perpendicular to each other.
Therefore, the equation of the conic can be written as
$4\times 5{\left(\frac{2y-x-3}{\sqrt{2^2+1^2}}\right)}^2-9\times 5{\left(\frac{2x+y-1}{\sqrt{2^2+1^2}}\right)}^2=80$
$\Rightarrow 4{\left(\frac{2y-x-3}{\sqrt{5}}\right)}^2-9{\left(\frac{2x+y-1}{\sqrt{5}}\right)}^2=16$

On putting $\frac{2y-x-3}{\sqrt{5}}=X$ and $\frac{2x+y-1}{\sqrt{5}}=Y,$ the given equation can be written as
$4X^2-9Y^2=16$
$\Rightarrow \frac{X^2}{4}-\frac{Y^2}{(4/3{)}^2}=1$
This is a Hyperbola.
Therefore, eccentricity is given by,
$e=\sqrt{1+\frac{16/9}{4}}$
$=\frac{\sqrt{13}}{3}$