Mathematics · 2022
JEE Main · 28 June 2022, Shift 2 · Q13
Let a >0, b >0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola x^2/(a^2)-y^2/(b^2)=1. Let e^′ and l^′…
Let $\displaystyle \mathrm{a}>0, \mathrm{~b}>0$. Let e and $\displaystyle l$ respectively be the eccentricity and length of the latus rectum of the hyperbola $\displaystyle \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$. Let $\displaystyle \mathrm{e}^{\prime}$ and $\displaystyle l^{\prime}$ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $\displaystyle \mathrm{e}^2=\frac{11}{14} l$ and $\displaystyle \left(\mathrm{e}^{\prime}\right)^2=\frac{11}{8} l^{\prime}$, then the value of $\displaystyle 77 \mathrm{a}+44 \mathrm{~b}$ is equal to :
Official answer
(4)
$\displaystyle 130$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.