A hyperbola passes through a focus of the ell | vedclass

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.The given equation of the ellipse is $\frac{x^2}{13^2} + \frac{y^2}{5

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$\frac{x^2}{144} - \frac{y^2}{25} = 1$

Vedclass · Answer

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.The given equation of the ellipse is $\frac{x^2}{13^2} + \frac{y^2}{5^2} = 1$.For the ellipse,$a = 13$ and $b = 5$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.The focus of the ellipse is $(\pm ae, 0) = (\pm 13 	imes \frac{12}{13}, 0) = (\pm 12, 0)$.Since the hyperbola passes through $(\pm 12, 0)$,we have $\frac{144}{a^2} - 0 = 1$,which implies $a^2 = 144$.The eccentricity of the hyperbola is $e' = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{b^2}{144}}$.Given that the product of eccentricities is $1$,we have $e 	imes e' = 1$.$\frac{12}{13} 	imes \sqrt{1 + \frac{b^2}{144}} = 1$.$\sqrt{1 + \frac{b^2}{144}} = \frac{13}{12}$.Squaring both sides,$1 + \frac{b^2}{144} = \frac{169}{144}$.$\frac{b^2}{144} = \frac{169}{144} - 1 = \frac{25}{144}$.Thus,$b^2 = 25$.The equation of the hyperbola is $\frac{x^2}{144} - \frac{y^2}{25} = 1$.

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

$A$ hyperbola passes through a focus of the ellipse $\frac{x^2}{169} + \frac{y^2}{25} = 1$. Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of their eccentricities is $1$. Then,the equation of the hyperbola is:

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