Let $P, Q$ be two points on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ whose eccentric angles differ by a right angle. The tangents at $P$ and $Q$ meet at $R$. If the chord $PQ$ divides the line segment $CR$ in $m:n$, then find $m/n$ (where $C$ is the centre of ellipse).
Step-by-Step Solution
Key Concept: Collinearity of three points is verified using the determinant condition, which after expansion relates the ratios in which $S$ divides the segment.
Given points $P(a\cos\alpha, b\sin\alpha)$ and $Q(-a\sin\alpha, b\cos\alpha)$ on an ellipse, the tangents at these points are $\frac{x\cos\alpha}{a} + \frac{y\sin\alpha}{b} = 1$ and $\frac{-x\sin\alpha}{a} + \frac{y\cos\alpha}{b} = 1$ respectively. These tangents meet at $R(a(\cos\alpha - \sin\alpha), b(\sin\alpha + \cos\alpha))$. Point $S$ on line $CR$ dividing it in ratio $m:n$ is $S\left(\frac{ma(\cos\alpha-\sin\alpha)}{m+n}, \frac{mb(\sin\alpha+\cos\alpha)}{m+n}\right)$. For collinearity of $P, Q, S$, the determinant condition yields $mR_1 + mR_2 - (m+n)R_3 = 0$, which simplifies to $m = n$, giving $m:n = 1:1$.
Correct Answer: 1