Let the length of the latus rectum of an ellipse x^2a^2+y^2b^2=1 (ab) be 30.

Question

Let the length of the latus rectum of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-t^2+2t-\dfrac34$, and the value of $a^2+b^2$ is required, then $a^2+b^2$ is equal to:

Accepted Answer

The maximum of $f(t)=-t^2+2t-\dfrac34$ occurs at $t=1$: $f(1)=-1+2-\dfrac34=\dfrac14$. So $e=\dfrac14$, i.e. $e^2=\dfrac1{16}$.

Latus rectum: $\dfrac{2b^2}{a}=30\Rightarrow b^2=15a$.

$$e^2=1-\frac{b^2}{a^2}=\frac{1}{16} \;\Rightarrow\; \frac{b^2}{a^2}=\frac{15}{16} \;\Rightarrow\; \frac{15a}{a^2}=\frac{15}{16}\;\Rightarrow\;a=16.$$

$$b^2=15\times16=240.$$

$a^2+b^2 = 256+240=496.$

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