Let the eccentricity of an ellipse x^2a^2+y^2b^2=1 be reciprocal to that of the hyperbola …

Question

Let the eccentricity of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be reciprocal to that of the hyperbola $2x^2-2y^2=1$. If the ellipse intersects the hyperbola at right angles, then the square of the length of the latus rectum of the ellipse is \_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

Hyperbola $x^2-y^2=\tfrac12$ is rectangular: $e'=\sqrt2$, foci $(\pm1,0)$. So $e_{\text{ellipse}}=\dfrac1{\sqrt2}\Rightarrow\dfrac{b^2}{a^2}=\dfrac12\Rightarrow a^2=2b^2$.

Curves intersecting orthogonally along common axes are confocal, so the ellipse's focus is also $1$: $ae=1\Rightarrow a\cdot\dfrac1{\sqrt2}=1\Rightarrow a=\sqrt2\Rightarrow a^2=2,\ b^2=1$.

$$\text{Latus rectum}=\frac{2b^2}a=\frac2{\sqrt2}=\sqrt2\ \Longrightarrow\ (\text{L.R.})^2=2.$$

Solutions · 1