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

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Hyperbola

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 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

77 a + 44 b

is equal to

Answer: (D)

For the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

e = 1 + \frac{b ^{2}}{a ^{2}}

and

ℓ = \frac{2 b ^{2}}{a}

. From

e^{2} = \frac{11}{14} ℓ

, we get

1 + \frac{b ^{2}}{a ^{2}} = \frac{11}{7} \cdot \frac{b ^{2}}{a}

, giving

\frac{a ^{2} + b ^{2}}{a ^{2}} = \frac{11 b ^{2}}{7 a}

... (1). For the conjugate hyperbola:

e^{'} = 1 + \frac{a ^{2}}{b ^{2}}

and

ℓ^{'} = \frac{2 a ^{2}}{b}

. From

(e^{'})^{2} = \frac{11}{8} ℓ^{'}

, we get

\frac{a ^{2} + b ^{2}}{b ^{2}} = \frac{11 a ^{2}}{4 b}

... (2). Dividing (1) by (2):

\frac{b ^{2}}{a ^{2}} = \frac{4 b ^{3}}{7 a ^{3}}

, so

7 a = 4 b

. Substituting back into (2) and solving:

b = \frac{65}{44}

. Therefore,

77 a + 44 b = 11 (7 a + 4 b) = 11 (4 b + 4 b) = 88 b = 88 \cdot \frac{65}{44} = 130

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option