Let the ellipse E : x^2144+ y^2169=1 and the hyperbola H : x^216- y^2 ^2=-1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H , then the valu…

Question

Let the ellipse E : x^2144+ y^2169=1 and the hyperbola H : x^216- y^2 ^2=-1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H , then the value of 24 (e+L ) is :

Accepted Answer

Correct answer: D. (d) Ellipse E : x^2144+ y^2169=1 has a^2=169 ,b^2=144, so c_E= 25=5 with foci at (0 , 5 ). Hyperbola H : y^2x^2- x^216=1 has the same foci, so c_H^2=x^2+16=25, giving x^2=9. Thus H is y^29- x^216=1 with a=3 ,b=4 ,c=5. Eccentricity: e= 53. Latus rectum: L= 2b^2a= 323. Therefore 24 (e+L )=24( 53+ 323)=24 373=296

$Let the ellipse E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1 and the hyperbola H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H , then the value of 24 (e + L) is :$

Solution

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