Let $\frac{x²}{a²}$ +$\frac{y²}{b²}$ = 1 (a > b) be a given ellipse. Length of its Latus rectum is 10. If its eccentricity is the maximum value of the function φ(t) = $\frac{5}{12}$ + t − t², then a² + b² is equal to:
126
The length of the latus rectum is \(\dfrac{2b^2}{a}=10 \Rightarrow b^2 = 5a\).
To find the maximum of \(\phi(t)=\dfrac{5}{12}+t-t^2\), differentiate and set to zero: \(\phi'(t)=1-2t=0 \Rightarrow t=\dfrac12\), and since φ''(t) = −2 < 0, this is a maximum.
Maximum value: \(\phi(1/2)=\dfrac{5}{12}+\dfrac12-\dfrac14=\dfrac{5}{12}+\dfrac{6}{12}-\dfrac{3}{12}=\dfrac{8}{12}=\dfrac23\). So the eccentricity \(e=\dfrac23\).
Using \(e^2 = 1-\dfrac{b^2}{a^2}\): \(\dfrac49 = 1-\dfrac{b^2}{a^2} \Rightarrow \dfrac{b^2}{a^2}=\dfrac59 \Rightarrow b^2=\dfrac{5a^2}{9}\).
Equating the two expressions for b²: \(5a = \dfrac{5a^2}{9} \Rightarrow 9a = a^2 \Rightarrow a = 9\) (since a ≠ 0).
Then \(b^2 = 5a = 45\) and \(a^2 = 81\), so \(a^2+b^2 = 81+45 = 126\).
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