Question

Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola
$\frac{x^2}{a^2}−\frac{y^2}{b^2}=1$
Let e′ and l′ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If
$e^2=\frac{11}{14}l$ and $(e^′)^2=\frac{11}{8}l^′$
then the value of 77a + 44b is equal to :

• A. 100
• B. 110
• C. 120
• D. 130

Answer 1

Answer: (D)

130

Answer 2

The correct answer is (D) : 130
$H : \frac{x^2}{a^2}−\frac{y^2}{b^2}=1$
then
$e^2=\frac{11}{14}l$
(I be the length of LR)
$⇒a^2+b^2=\frac{11}{7}b^2a…(i)$
and
$e^{′^2}=\frac{11}{8}l^′$
(I′ be the length of LR of conjugate hyperbola)
$⇒a^2+b^2=\frac{11}{4}a^2b…(ii)$
By (i) and (ii)
7a = 4b
then by (i)
$\frac{16}{49}b^2+b^2=\frac{11}{7}b^2⋅\frac{4b}{7}$
⇒ 44b = 65 and 77a = 65
Therefore , 77a + 44b = 130