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