Question

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

• A. $\frac{3}{5}$
• B. $\frac{1}{2}$
• C. $\frac{2}{5}$
• D. none of these

Answer 1

Answer: (A)

$\frac{3}{5}$

$25x^2+16y^2-150x-175=0\Rightarrow 25x^2-150x+225+16y^2-175-225=0\Rightarrow 25{(x-3)}^2+16{\left(y\right)}^2=400\Rightarrow \frac{{25(x-3)}^2}{400}+\frac{16y^2}{400}=1\Rightarrow \frac{{(x-3)}^2}{16}+\frac{y^2}{25}=1$

it is inthe form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b>a)$

here eccentricity$\left(e\right)=$$\sqrt{1-\frac{a^2}{b^2}}$

$e=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

so A is the correct answer