Let the eccentricity of an ellipse x^2a^2+y^2b^2=1, ab, be 14.

Question

Let the eccentricity of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b$, be $\dfrac14$. If this ellipse passes through the point $\left(-4\sqrt{\dfrac25},3\right)$, then $a^2+b^2$ is equal to:

Accepted Answer

$b^2=a^2\left(1-\tfrac1{16}\right)=\dfrac{15a^2}{16}$.

Substituting the point: $\dfrac{32/5}{a^2}+\dfrac9{b^2}=1\Rightarrow\dfrac{32}{5a^2}+\dfrac{144}{15a^2}=1\Rightarrow\dfrac{16}{a^2}=1\Rightarrow a^2=16$.

$$b^2=\frac{15}{16}\cdot16=15,\qquad a^2+b^2=31.$$

Solutions · 1