Hyperbola Question 7
Question 7 - 2024 (31 Jan Shift 1)
Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ be $( \pm 5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}}-\frac{y^{2}}{a^{2} b^{2}}=1$ equals
Show Answer
Answer (51)
Solution
focii $\equiv( \pm 5,0) ; \frac{2 b^{2}}{a}=\sqrt{50}$
$a=5 \quad b^{2}=\frac{5 \sqrt{2} a}{2}$
$b^{2}=a^{2}\left(1-e^{2}\right)=\frac{5 \sqrt{2} a}{2}$
$\Rightarrow a\left(1-e^{2}\right)=\frac{5 \sqrt{2}}{2}$
$\Rightarrow \frac{5}{e}\left(1-e^{2}\right)=\frac{5 \sqrt{2}}{2}$
$\Rightarrow \sqrt{2}-\sqrt{2} e^{2}=e$
$\Rightarrow \sqrt{2} e^{2}+e-\sqrt{2}=0$
$\Rightarrow \sqrt{2} e^{2}+2 e-e-\sqrt{2}=0$
$\Rightarrow \sqrt{2} e(e+\sqrt{2})-1(1+\sqrt{2})=0$
$\Rightarrow(e+\sqrt{2})(\sqrt{2} e-1)=0$
$\therefore e \neq-\sqrt{2} ; e=\frac{1}{\sqrt{2}}$
$\frac{x^{2}}{b^{2}}-\frac{y^{2}}{a^{2} b^{2}}=1 \quad a=5 \sqrt{2}$
$b=5$
$a^{2} b^{2}=b^{2}\left(e _1^{2}-1\right) \Rightarrow e _1^{2}=51$
