The straight line $\frac{lx}{a} + \frac{my}{b} = n$ meet the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at in the points $P$ and $Q$. If $OP$ and $OQ$ are along a pair of semi-conjugate diameters, $O$ being the centre of the ellipse, then $\frac{l^2}{n^2} + \frac{m^2}{n^2}$ equals____.
Step-by-Step Solution
Key Concept: For semi-conjugate diameters of an ellipse, if points P and Q lie on the ellipse at perpendicular directions from center O, they satisfy the conjugate diameter property: if P is at parameter α and Q at parameter β with |α - β| = π/2, then the chord PQ equation combined with the conjugate diameter condition yields a constraint on the line coefficients.
Given $P$ and $Q$ are at angles $\alpha$ and $\beta$ with $|\alpha - \beta| = \frac{\pi}{2}$, the equation of line $PQ$ is $\frac{1}{a}\cos\frac{\alpha+\beta}{2} + b\sin\frac{\alpha+\beta}{2} = \cos\frac{\alpha-\beta}{2} = \frac{1}{\sqrt{2}}$. For this to coincide with the given line, the coefficients must match.
Correct Answer: 2