KCET · Maths · Ellipse
The eccentric angle of the point \((2, \sqrt{3})\) lying on \(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) is
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
Given equation of an ellipse is
\(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) and point \(P(2, \sqrt{3})\)
Let \(\theta\) be the eccentric angle.
The parametric coordinate of an ellipse is
\[
\left.\begin{array}{l}
x=4 \cos \theta \\
y=2 \sin \theta
\end{array}\right\} \quad \text{...(i)}
\]
Given that, eccentric angle at \(P\) is,
\[
\begin{aligned}
2 &=4 \cos \theta \Rightarrow \cos \theta=\frac{1}{2} \\
\sqrt{2} &=2 \sin \theta \Rightarrow \sin \theta=\frac{\sqrt{3}}{2}
\end{aligned}
\]
Hence,
\[
\theta=\frac{\pi}{3}
\]
\(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) and point \(P(2, \sqrt{3})\)
Let \(\theta\) be the eccentric angle.
The parametric coordinate of an ellipse is
\[
\left.\begin{array}{l}
x=4 \cos \theta \\
y=2 \sin \theta
\end{array}\right\} \quad \text{...(i)}
\]
Given that, eccentric angle at \(P\) is,
\[
\begin{aligned}
2 &=4 \cos \theta \Rightarrow \cos \theta=\frac{1}{2} \\
\sqrt{2} &=2 \sin \theta \Rightarrow \sin \theta=\frac{\sqrt{3}}{2}
\end{aligned}
\]
Hence,
\[
\theta=\frac{\pi}{3}
\]
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