Question

If the line \(lx + my + n = 0\) cuts the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{25} = 1\) in points whose eccentric angles differ by \(\frac{\pi}{2}\), then \(\frac{a^2 l^2 + b^2 m^2}{n^2}\) is equal to

• A. 1
• B. 2
• C. 4
• D. \(\frac{3}{2}\)

Hint:

Use parametric form of ellipse and orthogonal condition.

Answer 1

The Correct Option is B

Step 1: Formula / Definition}
\[ \text{Points: } (a\cos\theta, b\sin\theta) \text{ and } (a\cos(\theta+\pi/2), b\sin(\theta+\pi/2)) = (-a\sin\theta, b\cos\theta) \]
Step 2: Calculation / Simplification}
Line passes through both points:
\(l(a\cos\theta) + m(b\sin\theta) + n = 0 \Rightarrow la\cos\theta + mb\sin\theta = -n\)
\(l(-a\sin\theta) + m(b\cos\theta) + n = 0 \Rightarrow -la\sin\theta + mb\cos\theta = -n\)
Square and add:
\((la\cos\theta + mb\sin\theta)^2 + (-la\sin\theta + mb\cos\theta)^2 = n^2 + n^2\)
\(l^2a^2(\cos^2\theta + \sin^2\theta) + m^2b^2(\sin^2\theta + \cos^2\theta) = 2n^2\)
\(a^2 l^2 + b^2 m^2 = 2n^2 \Rightarrow \frac{a^2 l^2 + b^2 m^2}{n^2} = 2\)
Step 3: Final Answer
\[ 2 \]