Question

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

• A. $\frac{5}{3}$
• B. $\frac{4}{3}$
• C. $\frac{\sqrt{11}}{3}$
• D. $\frac{\sqrt{10}}{3}$

Answer 1

The correct option is D $\frac{\sqrt{10}}{3}$

Here, $2y-x-3$ and $4x+2y-1$ are perpendicular to each other.
Therefore, the equation of the conic can be written as
$4\times 5{\left(\frac{2y-x-3}{\sqrt{2^2+1^2}}\right)}^2-9\times 20{\left(\frac{4x+2y-1}{\sqrt{4^2+2^2}}\right)}^2=80\Rightarrow {\left(\frac{2y-x-3}{\sqrt{2^2+1^2}}\right)}^2-9{\left(\frac{4x+2y-1}{\sqrt{4^2+2^2}}\right)}^2=4$

Putting $\frac{2y-x-3}{\sqrt{5}}=X,\frac{4x+2y-1}{\sqrt{5}}=Y$
We get
$X^2-9Y^2=4\Rightarrow \frac{X^2}{2^2}-\frac{Y^2}{(2/3{)}^2}=1$
This is a Hyperbola.
Therefore, eccentricity is given by,
$e=\sqrt{1+\frac{4/9}{4}}\therefore e=\frac{\sqrt{10}}{3}$