Question

The eccentric angle of a point on the ellipse \(x^2+3y^2=6\) lying at a distance of \(2\) units from its centre is

• A. \(\dfrac{\pi}{6}\)
• B. \(\dfrac{\pi}{4}\)
• C. \(\dfrac{\pi}{3}\)
• D. \(\dfrac{\pi}{2}\)

Hint:

For an ellipse, use \(x=a\cos\theta\) and \(y=b\sin\theta\), where \(\theta\) is the eccentric angle.

Answer 1

The Correct Option is B

Step 1: Write the ellipse in standard form.
The given ellipse is \[ x^2+3y^2=6 \] Dividing by \(6\), we get \[ \frac{x^2}{6}+\frac{y^2}{2}=1 \] So, \[ a^2=6,\qquad b^2=2 \] Hence, \[ a=\sqrt{6},\qquad b=\sqrt{2} \]

Step 2: Use the eccentric angle form.
For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] a point can be written as \[ x=a\cos\theta,\qquad y=b\sin\theta \] Therefore, \[ x=\sqrt{6}\cos\theta,\qquad y=\sqrt{2}\sin\theta \]

Step 3: Use the distance condition.
The centre of the ellipse is the origin \((0,0)\).
The point lies at a distance \(2\) units from the centre, so \[ x^2+y^2=2^2 \] \[ x^2+y^2=4 \]
Substituting the parametric values, \[ (\sqrt{6}\cos\theta)^2+(\sqrt{2}\sin\theta)^2=4 \] \[ 6\cos^2\theta+2\sin^2\theta=4 \] Dividing by \(2\), \[ 3\cos^2\theta+\sin^2\theta=2 \]

Step 4: Solve for \(\theta\).
Using \[ \sin^2\theta=1-\cos^2\theta, \] we get \[ 3\cos^2\theta+1-\cos^2\theta=2 \] \[ 2\cos^2\theta=1 \] \[ \cos^2\theta=\frac{1}{2} \] Thus, \[ \cos\theta=\frac{1}{\sqrt{2}} \] Hence, \[ \theta=\frac{\pi}{4} \]

Step 5: Final conclusion.
Therefore, the eccentric angle is \[ \boxed{\frac{\pi}{4}} \]