Question

If a point \( P(x,y) \) moves along the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) and \( C \) is the centre of the ellipse, then \( 4\max\{CP\} + 5\min\{CP\} \) is

• A. 25
• B. 40
• C. 45
• D. 54
• E. 16

Hint:

For ellipse centered at origin, farthest point from center lies along major axis and closest along minor axis.

Answer 1

The Correct Option is B

Concept:
• The ellipse represents a set of points whose distance from center varies continuously.
• For ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):
• Maximum distance from center = semi-major axis \(a\)
• Minimum distance from center = semi-minor axis \(b\)
• This can also be derived analytically using distance formula.

Step 1: Identify ellipse parameters.
Given: \[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \] Comparing with standard form: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] We get: \[ a^2 = 25 \Rightarrow a = 5 \] \[ b^2 = 16 \Rightarrow b = 4 \]

Step 2: Locate center of ellipse.
The given ellipse is centered at origin: \[ C = (0,0) \]

Step 3: Express distance \(CP\).
Distance from center to any point \(P(x,y)\) is: \[ CP = \sqrt{x^2 + y^2} \] We need to find: \[ \max CP \quad \text{and} \quad \min CP \]

Step 4: Use parametric form of ellipse.
Let: \[ x = 5\cos\theta, \quad y = 4\sin\theta \] Substitute into distance: \[ CP = \sqrt{(5\cos\theta)^2 + (4\sin\theta)^2} \] \[ = \sqrt{25\cos^2\theta + 16\sin^2\theta} \]

Step 5: Simplify expression.
\[ CP^2 = 25\cos^2\theta + 16\sin^2\theta \] Rewrite: \[ = 25(1 - \sin^2\theta) + 16\sin^2\theta \] \[ = 25 - 25\sin^2\theta + 16\sin^2\theta \] \[ = 25 - 9\sin^2\theta \]

Step 6: Find maximum and minimum values.
Since: \[ 0 \le \sin^2\theta \le 1 \] Maximum of \(CP^2\): \[ = 25 - 9(0) = 25 \Rightarrow CP = 5 \] Minimum of \(CP^2\): \[ = 25 - 9(1) = 16 \Rightarrow CP = 4 \]

Step 7: Substitute into expression.
\[ 4 \times \max(CP) + 5 \times \min(CP) \] \[ = 4(5) + 5(4) \] \[ = 20 + 20 = 40 \]

Step 8: Final Answer.
\[ \boxed{40} \]