Question

The eccentricity of the ellipse represented by the equation $25x^2+16y^2-150x-175=0$ is

• A. $\frac{2}{5}$
• B. $\frac{3}{5}$
• C. $\frac{4}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{3}{5}$

$25x^2+16y^2-150x-175=0$
$\Rightarrow 25(x^2-6x+9)+16y^2=175+25\times 9$
$\Rightarrow 25(x-3{)}^2+16y^2=400$
$\Rightarrow \frac{(x-3{)}^2}{16}+\frac{y^2}{25}=1$
$a=4,b=5(b>a)$
$\therefore a^2=b^2(1-e^2)$
$\Rightarrow 16=25(1-e^2)$
$\Rightarrow e^2=1-\frac{16}{25}$
$\Rightarrow e=\frac{3}{5}$