Let e_{1} and e_{2} be the eccentricities of | vedclass

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(A) For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$ $(b < 5)$,the eccentricity $e_{1}$ satisfies $b^{2}=25(1-e_{1}^{2})$.
For the hyperbo

Vedclass · Accepted Answer

$(8, 10)$

Vedclass · Answer

(A) For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$ $(b < 5)$,the eccentricity $e_{1}$ satisfies $b^{2}=25(1-e_{1}^{2})$.
For the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$,the eccentricity $e_{2}$ satisfies $b^{2}=16(e_{2}^{2}-1)$.
Equating the two expressions for $b^{2}$,we get $25(1-e_{1}^{2})=16(e_{2}^{2}-1)$.
Given $e_{1}e_{2}=1$,we substitute $e_{2}=\frac{1}{e_{1}}$:
$25(1-e_{1}^{2})=16(\frac{1}{e_{1}^{2}}-1) = 16(\frac{1-e_{1}^{2}}{e_{1}^{2}})$.
Since $b  < 5$,$e_{1} 
eq 1$,so we divide by $(1-e_{1}^{2})$ to get $25 = \frac{16}{e_{1}^{2}}$,which implies $e_{1}^{2}=\frac{16}{25}$,so $e_{1}=\frac{4}{5}$.
Then $e_{2}=\frac{1}{e_{1}}=\frac{5}{4}$.
The distance between the foci of the ellipse is $2ae_{1} = 2(5)(\frac{4}{5}) = 8 = \alpha$.
The distance between the foci of the hyperbola is $2ae_{2} = 2(4)(\frac{5}{4}) = 10 = \beta$.
Thus,the ordered pair $(\alpha, \beta) = (8, 10)$.

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

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