Let a0, b0.

Question

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

Accepted Answer

$$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, then

$${e^2} = {{11} \over {14}}l$$ (l be the length of LR)

$$ \Rightarrow {a^2} + {b^2} = {{11} \over 7}{b^2}a$$ ..... (i)

and $$e{'^2} = {{11} \over 8}l'$$ (l' be the length of LR of conjugate hyperbola)

$$ \Rightarrow {a^2} + {b^2} = {{11} \over 4}{a^2}b$$ ....... (ii)

By (i) and (ii)

$7a = 4b$

then by (i)

$${{16} \over {49}}{b^2} + {b^2} = {{11} \over 7}{b^2}\,.\,{{4b} \over 7}$$

$$ \Rightarrow 44b = 65$$ and $77a = 65$

$$\therefore$$ $77a + 44b = 130$

Solutions · 1