Q97 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

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

  • A
    -3
  • B
    3
  • Correct answer
    C
    -9
  • D
    9

Solution

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.

Hence, the correct answer is option: 'C'.

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