Let the ellipse E: (x^2)/(144) + (y^2)/(169) = 1 and the hyperbola H: (x^2)/(16) - (y^2)/(…

Question

Let the ellipse $E: \frac{x^2}{144} + \frac{y^2}{169} = 1$ and the hyperbola $H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$ have the same foci. If $e$ and $L$ 
respectively denote the eccentricity and the length of the latus rectum of $H$...

Accepted Answer

The equation of the ellipse is :

$E:\frac{{x}^{2}}{144}+\frac{{y}^{2}}{169}=1$

Comparing this with the standard form $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$, we have :

${a}^{2}=144\Rightarrow a=12$

${b}^{2}=169\Rightarrow b=13$

Since $b>a$, the ellipse is vertically oriented. The distance from the center to the foci (c) is given by :

$\begin{gathered} {c}^{2}={b}^{2}-{a}^{2} \ \, {c}^{2}=169-144=25\implies c=5 \end{gathered}$

The foci of the ellipse are at $(0,\pm 5)$.

The equation of the hyperbola is :

$H:\frac{{x}^{2}}{16}-\frac{{y}^{2}}{{\lambda }^{2}}=-1\,$ which is equivalent to $\, \frac{{y}^{2}}{{\lambda }^{2}}-\frac{{x}^{2}}{16}=1$

This is a vertical hyperbola.

Let ${B}^{2}={\lambda }^{2}$ and ${A}^{2}=16$.

Since it shares the same foci as the ellipse, its focal distance $c$ is also 5 . For a hyperbola:

$\begin{gathered} {c}^{2}={A}^{2}+{B}^{2} \ 25=16+{\lambda }^{2}\implies {\lambda }^{2}=9\implies \lambda =3 \end{gathered}$

Calculate Eccentricity ( $e$ ) and Latus Rectum ( $L$ ) of $H$

Eccentricity (e):

$e=\frac{c}{B}=\frac{5}{3}$

Length of Latus Rectum ( $L$ ): For a vertical hyperbola $\frac{{y}^{2}}{{B}^{2}}-\frac{{x}^{2}}{{A}^{2}}=1$, the length of the latus rectum is:

$L=\frac{2{A}^{2}}{B}=\frac{2	imes 16}{3}=\frac{32}{3}$

We need to find $24(e+L)$ :

$\begin{gathered} e+L=\frac{5}{3}+\frac{32}{3}=\frac{37}{3} \ 	herefore 24(e+L)=24	imes \frac{37}{3}=8	imes 37=296 \end{gathered}$

Solutions · 1