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 values of $\cot \frac{\alpha}{2} . \cot \frac{\beta}{2}$ is

• A. -3
• B. 3
• C. -9
• D. 9

Answer 1

Answer: (C)

-9

Answer 2

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 $(4,0)$
$\frac{4}{5} \cos \frac{\alpha+\beta}{2}=\cos \frac{\alpha-\beta}{2}$
$\Rightarrow 4\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right)$
$=5\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}\right)$
$\Rightarrow 4\left(\cot \frac{\alpha}{2} \cot \frac{\beta}{2}-1\right)$
$=5\left(\cot \frac{\alpha}{2} \cot \frac{\beta}{2}+1\right)$
$\Rightarrow \cot \frac{\alpha}{2} \cot \frac{\beta}{2}=-9$