JEE Main 2022 (Online) 28th June Evening ShiftQuestion 1
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 :
Hyperbola
Solution
Let $r=\dfrac ba$. $e^2=1+r^2$, $l=\dfrac{2b^2}{a}=2r^2a$; $e'^2=1+\dfrac1{r^2}$, $l'=\dfrac{2a^2}{b}=\dfrac{2a}{r}$. $1+r^2=\dfrac{11}{14}\cdot2r^2a$ and $1+\dfrac{1}{r^2}=\dfrac{11}{8}\cdot\dfrac{2a}{r}$. The second gives $a=\dfrac{4(1+r^2)}{11r}$. Substitute in the first: $1+r^2=\dfrac47r(1+r^2)\Rightarrow r=\dfrac74$, so $a=\dfrac{65}{77}$, $b=\dfrac{65}{44}$. $77a+44b=65+65=130$.