JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let the length of a latus rectum of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) be 10 . If its eccentricity is the minimum value of the function \(f(\mathrm{t})=\mathrm{t}^2+\mathrm{t}+\frac{11}{12}\), \(\mathrm{t} \in \mathbf{R}\), then \(\mathrm{a}^2+\mathrm{b}^2\) is equal to :
- A 125
- B 126
- C 120
- D 115
Answer & Solution
Correct Answer
(B) 126
Step-by-step Solution
Detailed explanation
Length of LR \(=\frac{2 b^2}{a}=10 \Rightarrow 5 \mathrm{a}=\mathrm{b}^2\)...(1)…
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