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(B) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord,then t

Vedclass · Accepted Answer

$-9$

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(B) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord,then the relation is $	an(\alpha/2) 	an(\beta/2) = -\frac{1-e}{1+e}$ or $	an(\alpha/2) 	an(\beta/2) = -\frac{b}{a}$.Given $\frac{x^2}{25} + \frac{y^2}{9} = 1$,we have $a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.The condition for a focal chord is $	an(\alpha/2) 	an(\beta/2) = -\frac{b}{a} = -\frac{3}{5}$.We need to find $\frac{\cot(\alpha/2)}{	an(\beta/2)} = \frac{1}{	an(\alpha/2) 	an(\beta/2)}$.Substituting the value,we get $\frac{1}{-3/5} = -\frac{5}{3}$.Note: The standard property for focal chords is $	an(\alpha/2) 	an(\beta/2) = \frac{e-1}{e+1}$. For the given ellipse,$e = \sqrt{1 - 9/25} = 4/5$.Thus,$	an(\alpha/2) 	an(\beta/2) = \frac{4/5 - 1}{4/5 + 1} = \frac{-1/5}{9/5} = -1/9$.Therefore,$\frac{\cot(\alpha/2)}{	an(\beta/2)} = \frac{1}{	an(\alpha/2) 	an(\beta/2)} = -9$.

If the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ are $\alpha$ and $\beta$,then $\frac{\cot(\alpha/2)}{\tan(\beta/2)}=$

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