For the hyperbola H:x^2-y^2=1 and the ellipse E:x^2a^2+y^2b^2=1, ab0, let (1) the eccent…

Question

For the hyperbola $H:x^2-y^2=1$ and the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b>0$, let (1) the eccentricity of $E$ be reciprocal of the eccentricity of $H$, and (2) the line $y=\sqrt{\dfrac52}\,x+K$ be a common tangent of $E$ and $H$. Then $4(a^2+b^2)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$e_H=\sqrt2\Rightarrow e_E=\dfrac1{\sqrt2}\Rightarrow1-\dfrac{b^2}{a^2}=\dfrac12\Rightarrow a^2=2b^2$.

Equal tangent lines require $a^2m^2+b^2=m^2-1$ for both curves at slope $m^2=\dfrac52$:
$$5(1-a^2)=2(1+b^2)\ \Longrightarrow\ 5a^2+2b^2=3.$$

With $a^2=2b^2$: $12b^2=3\Rightarrow b^2=\dfrac14,\ a^2=\dfrac12$.

$$4(a^2+b^2)=4\cdot\frac34=3.$$

Solutions · 1