JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the foci of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{7}=1\) and the hyperbola \(\frac{ x ^{2}}{144}-\frac{ y ^{2}}{\alpha}=\frac{1}{25}\) coincide. Then the length of the latus rectum of the hyperbola is:-
- A \(\frac{32}{9}\)
- B \(\frac{18}{5}\)
- C \(\frac{27}{4}\)
- D \(\frac{27}{10}\)
Answer & Solution
Correct Answer
(D) \(\frac{27}{10}\)
Step-by-step Solution
Detailed explanation
Ellipse : \(\frac{x^{2}}{16}+\frac{y^{2}}{7}=1\) Eccentricity \(=\sqrt{1-\frac{7}{16}}=\frac{3}{4}\) Foci \(\equiv(\pm a \quad e, 0) \equiv(\pm 3,0)\) Hyperbola : \(\frac{x^{2}}{\left(\frac{144}{25}\right)}-\frac{y^{2}}{\left(\frac{\alpha}{25}\right)}=1\) Eccentricity…
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