Question

Eccentricity of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, if it passes through point $(9,5)$ and $(12,4)$ is

• A. $\sqrt{\frac{3}{4}}$
• B. $\sqrt{\frac{4}{5}}$
• C. $\sqrt{\frac{5}{6}}$
• D. $\sqrt{\frac{6}{7}}$

Answer 1

The correct option is D $\sqrt{\frac{6}{7}}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through (9,5) and (12,4)

Therefore, $\frac{81}{a^2}+\frac{25}{b^2}=1$

$81b^2+25a^2=a^2{b}^2⟶\left(1\right)$

and

$\frac{144}{a^2}+\frac{16}{b^2}=1$

$144b^2+16a^2=a^2{b}^2$ $⟶\left(2\right)$

$eqn\left(2\right)=eqn\left(1\right)$

$81b^2+25a^2={144b}^2+16a^2$

${9a}^2={63b}^2$

$\frac{b^2}{a^2}=\frac{9}{63}$

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{63}}=\sqrt{\frac{54}{63}}$

$e=\frac{3}{3}\sqrt{\frac{6}{7}}$

$e=\sqrt{\frac{6}{7}}$