Question

If \(θ\) is the eccentric angle of a point on the ellipse \(\frac{x^2}{25}+\frac{y^2}{9} = 1\) such that the distance of the point from the center is \(5\), then \(θ =\).......

• A. \(0\)
• B. \(\frac{π}{6}\)
• C. \(\frac{π}{3}\)
• D. \(\frac{π}{2}\)

Hint:

Use P = (a cos t, b sin t) and set OP squared equal to 25.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A point on the ellipse \(\frac{x^2}{25} + \frac{y^2}{9} = 1\) with eccentric angle \(\theta\) is \((5\cos\theta,\ 3\sin\theta)\).

Step 2: Use the distance condition:
\[ OP^2 = 25\cos^2\theta + 9\sin^2\theta = 25 \]
\[ 25\cos^2\theta + 9(1 - \cos^2\theta) = 25 \Rightarrow 16\cos^2\theta = 16 \]
So \(\cos^2\theta = 1\), which gives \(\theta = 0\) (or \(\pi\)).

Step 3: Interpretation:
The distance from the centre is 5 only at the end of the major axis, \((\pm5, 0)\), where the semi-major axis \(a = 5\).

Step 4: Why the other options are wrong.
At \(\theta = \frac\pi6, \frac\pi3, \frac\pi2\) the distance \(OP\) is less than 5, for example 3 at \(\theta = \frac\pi2\).

Final Answer:
The eccentric angle is \(0\), option (A). \[ \boxed{0} \]