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

Question

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, -3, 3, -9, 9

MARKS App · Accepted Answer

Since equation of chord joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ is $\frac{x}{5} \cos \frac{\alpha+\beta}{2}+\frac{y}{3} \sin \frac{\alpha+\beta}{2}=\cos \frac{\alpha-\beta}{2}$ ...(i) $\because$ Chord (i) is the focal chord so, it will pass through focus $(4,0)$ $\begin{aligned} & \frac{4}{5} \cos \frac{\alpha+\beta}{2}=\cos \frac{\alpha-\beta}{2} \ & \Rightarrow 4\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right) \ & =5\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right) \ & \Rightarrow 4\left(\cot \frac{\alpha}{2} \cot \frac{\beta}{2}-1\right)=5\left(\cot \frac{\alpha}{2} \cot \frac{\beta}{2}+1\right) \ & \Rightarrow \cot \frac{\alpha}{2} \cot \frac{\beta}{2}=-9 \end{aligned}$ Hence, option (3) is correct.

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