Q53 / 110 · VITEEE 2023 · Mathematics

If the eccentricity of the hyperbola \(x^2-y^2 \sec ^2 \theta=4\) is \(\sqrt{3}\) times the eccentricity of the ellipse \(x^2 \sec ^2 \theta+y^2=16\), then the value of \(\theta\) equals

  • A
    \(\frac{\pi}{6}\)
  • Correct answer
    B
    \(\frac{3 \pi}{4}\)
  • C
    \(\frac{\pi}{3}\)
  • D
    \(\frac{\pi}{2}\)

Solution

Given: \(x^2-y^2 \sec ^2 \theta\) \(=4\) and \(x^2-\sec ^2 \theta+y^2=16\) \(\Rightarrow \frac{x^2}{4}-\frac{y^2}{4 \cos ^2 \theta}=1\) and \(\frac{x^2}{16 \cos ^2 \theta}+\frac{y^2}{16}=1\) According to problem \(\begin{aligned} & \frac{4+4 \cos ^2 \theta}{4}=3\left(\frac{16-16 \cos ^2 \theta}{16}\right) \\ & \Rightarrow 1+\cos ^2 \theta=3\left(1-\cos ^2 \theta\right) \Rightarrow 4 \cos ^2 \theta=2 \\ & \Rightarrow \cos \theta= \pm \frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}, \frac{3 \pi}{4} \end{aligned}\)

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

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