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(C) The equation of the chord joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by $\frac{x}{a

Vedclass · Accepted Answer

-$9$

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(C) The equation of the chord joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by $\frac{x}{a} \cos \frac{\alpha+\beta}{2} + \frac{y}{b} \sin \frac{\alpha+\beta}{2} = \cos \frac{\alpha-\beta}{2}$.For the given ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$,we have $a=5$ and $b=3$. The focus is $(ae, 0) = (5 \cdot \frac{4}{5}, 0) = (4, 0)$.Since the chord passes through the focus $(4, 0)$,we substitute $x=4$ and $y=0$ into the chord equation:$\frac{4}{5} \cos \frac{\alpha+\beta}{2} = \cos \frac{\alpha-\beta}{2}$.Using the expansion $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$:$4(\cos \frac{\alpha}{2} \cos \frac{\beta}{2} - \sin \frac{\alpha}{2} \sin \frac{\beta}{2}) = 5(\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2})$.Dividing both sides by $\sin \frac{\alpha}{2} \sin \frac{\beta}{2}$ (assuming $\sin \frac{\alpha}{2}, \sin \frac{\beta}{2} 
eq 0$):$4(\cot \frac{\alpha}{2} \cot \frac{\beta}{2} - 1) = 5(\cot \frac{\alpha}{2} \cot \frac{\beta}{2} + 1)$.$4 \cot \frac{\alpha}{2} \cot \frac{\beta}{2} - 4 = 5 \cot \frac{\alpha}{2} \cot \frac{\beta}{2} + 5$.$-\cot \frac{\alpha}{2} \cot \frac{\beta}{2} = 9$.$\cot \frac{\alpha}{2} \cot \frac{\beta}{2} = -9$.

If the line joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ is a focal chord,then one possible value of $\cot \frac{\alpha}{2} \cdot \cot \frac{\beta}{2}$ is

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