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: Geometric thought:
The farthest point of the ellipse from its centre is the end of the major axis, at distance $a = 5$. Any other point is closer.

Step 2: So the point is (5, 0):
The point $(5, 0)$ corresponds to $5\cos\theta = 5$, $3\sin\theta = 0$, so $\theta = 0$.

Step 3: Confirm algebraically:
$OP^2 = 9 + 16\cos^2\theta$; setting this to 25 gives $\cos^2\theta = 1$.

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