Let the foci of a hyperbola coincide with the foci of the ellipse (x^2)/(36)+(y^2)/(16)=1.

Question

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is :

Accepted Answer

$\begin{gathered} E:\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1 \ H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1 \end{gathered}$

For con-focal $E$ and $H$

$\because \, {A}^{2}-{B}^{2}={a}^{2}+{b}^{2}$

(Given that ${A}^{2}=36$ and ${B}^{2}=16$ )

$\begin{gathered} 36-16={a}^{2}+{b}^{2} \ {a}^{2}+{b}^{2}=20\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i) \ {e}_{H}=5 \ \Rightarrow \, 25=1+\frac{{b}^{2}}{{a}^{2}} \ \Rightarrow \, 25{a}^{2}=20\Rightarrow {a}^{2}=\frac{4}{5}\Rightarrow a=\frac{2}{\sqrt{5}} \end{gathered}$

Also ${a}^{2}+{b}^{2}=20$

$\begin{gathered} {b}^{2}=20-\frac{4}{5}\Rightarrow \frac{96}{5} \ L(LR)==2	imes \frac{{b}^{2}}{a}=\frac{2	imes \frac{96}{5}}{\frac{2}{\sqrt{5}}}=\frac{96}{\sqrt{5}} \end{gathered}$

Solutions · 1