Equation of the ellipse whose axes are the axes of coordinates and which passes through th…

Question

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1)$ and has eccentricity $\sqrt{2/5}$ is:

Accepted Answer

Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. With $e^2=\tfrac25$: $b^2=a^2(1-e^2)=\tfrac{3a^2}{5}\Rightarrow5b^2=3a^2$.

Through $(-3,1)$: $\dfrac{9}{a^2}+\dfrac{1}{b^2}=1$. Substituting $b^2=\tfrac{3a^2}{5}$:
$$\frac9{a^2}+\frac{5}{3a^2}=1\Rightarrow\frac{27+5}{3a^2}=1\Rightarrow a^2=\frac{32}{3},\quad b^2=\frac{32}{5}.$$
Equation: $\dfrac{x^2}{32/3}+\dfrac{y^2}{32/5}=1\Rightarrow3x^2+5y^2=32.$

Solutions · 1