Question

The eccentricity of the ellipse $9x^2+25y^2-18x-100y-116=0,$ is

• A. 25/16
  
• B. 4/5
  
• C. 16/25
  
• D. 5/4

Answer 1

The correct option is D 5/4

$9x^2+25y^2-18x-100y-116=0$
$\Rightarrow 9(x^2-2x)+25(y^2-4y)=116$
$\Rightarrow 9(x^2-2x+1)+25(y^2-4y+4)=116+100+9$
$\Rightarrow 9(x^2-1{)}^2+25(y-2{)}^2=225$
$\Rightarrow \frac{9(x-1{)}^2}{225}+\frac{25(y-2{)}^2}{225}=1$
$\Rightarrow \frac{(x-1{)}^2}{25}+\frac{(y-2{)}^2}{9}=1$
Comparing it with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get:
a=5 and b=3
Here, a>b so the major and the minor axes of the ellipse are along the x-axis and y-axis, respectively
Now, $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{9}{25}}$
$\Rightarrow e=\sqrt{\frac{16}{25}}$
$\Rightarrow e=\frac{4}{5}$