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JEE Mains · Maths · STD 12 - 6. Application of derivatives
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 \(126\)
- B \(135\)
- C \(145\)
- D \(116\)
Answer & Solution
Correct Answer
(A) \(126\)
Step-by-step Solution
Detailed explanation
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b) ; \frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a \ldots(i)\) \(Now , \quad \phi( t )=\frac{5}{12}+ t - t ^{2}=\frac{8}{12}-\left( t -\frac{1}{2}\right)^{2}\)…
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