Question

Let \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a>b)\) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, \(\phi(t)=\frac{5}{12}+t-t^2\), then \(a^2+b^2\) is equal to :

• A. 116
• B. 126
• C. 135
• D. 145

Hint:

Maximum of \(\phi\) is at \(t=\frac12\), giving \(e=\frac23\); then use \(\frac{2b^2}{a}=10\) and \(b^2=a^2(1-e^2)\).

Answer 1

The Correct Option is B

\(\frac{d\phi}{dt}=1-2t=0\Rightarrow t=\frac12\); \(\phi_{max}=\frac{5}{12}+\frac12-\frac14=\frac23\). So \(e=\frac23\).
Latus rectum \(\frac{2b^2}{a}=10\Rightarrow b^2=5a\). Also \(b^2=a^2(1-e^2)=\frac{5a^2}{9}\).
\(5a=\frac{5a^2}{9}\Rightarrow a=9\), \(b^2=45\).
\(a^2+b^2=81+45=126\).
Answer: (2) 126