If e1 and e2 are the eccentricities of the ellipse, \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1 \textrm{ }\textrm{ } and \textrm{ }\textrm{ } the \textrm{ }\textrm{ } hyperbola , \textrm{ }\textrm{ } \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1 respectively and (e1, e2) is a point on the ellipse, 15x2 + 3y2 = k, then k is equal to

Engineering

Mathematics

Standard Hyperbola

Question

If e₁and e₂ are the eccentricities of the ellipse, $\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1 and the hyperbola, \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ respectively and (e₁, e₂) is a point on the ellipse, 15x² + 3y² = k, then k is equal to

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$e_{1} = \sqrt{1 - \frac{4}{18}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$

$e_{2} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$

$15 e_{1}^{2} + 3 e_{2}^{2} = k \Rightarrow k = 15 (\frac{7}{9}) + 3 (\frac{13}{9}) ∴ k = 16$