Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1508 Upvotes
Verified Answer
The correct answer is:
\(\frac{3 \pi}{4}\)
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
\(\begin{aligned}
& \frac{4+4 \cos ^2 \theta}{4}=3\left(\frac{16-16 \cos ^2 \theta}{16}\right) \\
& \Rightarrow 1+\cos ^2 \theta=3\left(1-\cos ^2 \theta\right) \Rightarrow 4 \cos ^2 \theta=2 \\
& \Rightarrow \cos \theta= \pm \frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}, \frac{3 \pi}{4}
\end{aligned}\)
\(=4\) 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
\(\begin{aligned}
& \frac{4+4 \cos ^2 \theta}{4}=3\left(\frac{16-16 \cos ^2 \theta}{16}\right) \\
& \Rightarrow 1+\cos ^2 \theta=3\left(1-\cos ^2 \theta\right) \Rightarrow 4 \cos ^2 \theta=2 \\
& \Rightarrow \cos \theta= \pm \frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}, \frac{3 \pi}{4}
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.