JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If \(\mathrm{e}_{1}\) and \(\mathrm{e}_{2}\) are the eccentricities of the ellipse, \(\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{4}=1\) and the hyperbola, \(\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1\) respectively and \(\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)\) is a point on the ellipse, \(15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k},\) then \(\mathrm{k}\) is equal to
- A \(15\)
- B \(14\)
- C \(17\)
- D \(16\)
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
\(e_{1}=\sqrt{1-\frac{4}{18}}=\frac{\sqrt{7}}{3}\) \(\mathrm{e}_{2}=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}\) \(\because \quad\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)\) lies on \(15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k}\)…
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