Step-by-Step Solution
Key Concept: For ellipses, eccentricity is calculated as e = √(1 - b²/a²) where a > b, and the sum of distances from any point to the two foci equals 2a. For normals to conics, use the condition that the normal at parameter θ has specific slope relationships and passes through defined points on the curve.
For part (B), the ellipse equation $\frac{(x+1)^2}{9} + \frac{(y+2)^2}{25} = 1$ has $e^2 = 1 - \frac{9}{25} = \frac{16}{25}$, so $e = \frac{4}{5}$. The foci are at $S(-1, -2+4) = S(-1, 2)$ and $S'(-1, -6)$, giving sum of distances = $2 + 6 = 8$. For part (C), the normal at $(3\cos\theta, 2\sin\theta)$ is $3x\sec\theta - 2y\csc\theta = 5$, which is parallel to $2x + y = 1$ when $\cos\theta = -\frac{3}{5}$ and $\sin\theta = \pm\frac{4}{5}$, yielding points $(\frac{-9}{5}, \pm\frac{8}{5})$ with sum of distances $\frac{16}{5}$. For part (D), any point on line $x - y + 2 = 0$ gives chord of contact $(4y-2) + t(x+2y) = 0$, which passes through the intersection point $(-1, \frac{1}{2})$ at distance 3 from $(2, \frac{1}{2})$.
Correct Answer: [A-p,q,r,s] [B-p,q,r,s] [C-q,r,s] [D-p]