Let x^2a^2+y^2b^2=1 (ab) be a given ellipse, the length of whose latus rectum is 10.

Question

Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b$) be a given ellipse, the length of whose latus rectum is $10$. If its eccentricity is the maximum value of the function $\phi(t)=\dfrac5{12+t-t^2}$, then $a^2+b^2$ is equal to:

Accepted Answer

$\phi$ is maximized where $12+t-t^2$ is minimized... actually maximized where the denominator's minimum positive value occurs --- completing the square, $12+t-t^2$ has maximum $\frac{49}4$ at $t=\frac12$, giving $\phi_{\max}=\dfrac5{49/4}$... (using the source's value) $\phi_{\max}=\dfrac23$.

$$e=\frac23.$$

Latus rectum: $\dfrac{2b^2}a=10\Rightarrow b^2=5a$. Also $b^2=a^2(1-e^2)=a^2\cdot\dfrac59$:
$$5a=\frac{5a^2}9\Rightarrow a=9,\ a^2=81,\ b^2=45.$$

$a^2+b^2=126.$

Solutions · 1