Question

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

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

Answer 1

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

To find eccentricity of the conic
$4(2y-x-3{)}^2-9(2x+y-1{)}^2=80$
Here , $2y-x-3=0$ and $2x+y-1=0$ are perpendicular to each other
Therefore , the equation of the conic can be written as
$4\times 5\left[\frac{(2y-x-3{)}^2}{(\sqrt{(2{)}^2+(-1{)}^2}{)}^2}\right]-9\times 5\left[\frac{(2x+y-1{)}^2}{(\sqrt{(2{)}^2+(-1{)}^2}{)}^2}\right]=80$
$4\left[\frac{(2y-x-3{)}^2}{(\sqrt{5}{)}^2}\right]-9\left[\frac{(2x+y-1{)}^2}{(\sqrt{5}{)}^2}\right]=16$
Let , $X=\frac{2y-x-3}{\sqrt{5}}$ and
$Y=\frac{2x+y-1}{\sqrt{5}}$

The above equation becomes ,
$4X^2-9Y^2=16$
The above equation can be written as
$\frac{X^2}{4^2}-\frac{Y^2}{{\left(\frac{4}{3}\right)}^2}=1$
Which represents a hyperbola
Here , $a^2=4$ and $b^2=\frac{16}{9}$
$b^2=a^2(e^2-1)$
$(e^2-1)=\frac{4}{9}$
$e^2=\frac{13}{9}$
$e=\frac{\sqrt{13}}{3}$
Hence , Option B