If $e_1$ and $e_2$ are the roots of the equation $x^2 - ax + 2 = 0$, then match the following.
(A) If $e_1$ and $e_2$ are the eccentricities of the ellipse and hyperbola, respectively then the values of $a$ are
(B) If both $e_1$ and $e_2$ are the eccentricities of the hyperbolas, then values of $a$ are
(C) If $e_1$ and $e_2$ are eccentricities of hyperbola and conjugate hyperbola, then values of $a$ are
(D) If $e_1$ is the eccentricity of the hyperbola for which no such points exist from which perpendicular tangents can be drawn, then the values of $a$ are
Step-by-Step Solution
Key Concept: For an ellipse, eccentricity $e \in (0,1)$; for a hyperbola, $e > 1$. For conjugate hyperbolas with eccentricities $e_1$ and $e_2$, the relation $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$ holds. Using Vieta's formulas ($e_1 + e_2 = a$, $e_1 e_2 = 2$) and these constraints on eccentricity ranges determines the valid values of parameter $a$.
The problem requires finding parameter $a$ such that three conditions hold: (i) $D ≥ 0$ gives $a ∈ (-∞, -2\sqrt{2}] ∪ [2\sqrt{2}, ∞)$, (ii) $f(1) > 0$ or $1-a+2 > 0$ gives $a 1$ gives $a > 2$. Combining these three conditions yields $a ∈ (2\sqrt{2}, 3)$. For option (C), $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$ leads to $a = ±2\sqrt{2}$. For option (D), since $\sqrt{2} 2\sqrt{2}$.
Correct Answer: [A-p, s] [B-q] [C-r] [D-p, q, s]