Consider the quadratic equation $x^{2} - 8 p x + p - 2 = 0$ with roots $e_{1}$ and $e_{2}$ then:

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if $e_{1}$ and $e_{2}$ are eccentricities of two ellipses then $p \in (- \infty, \frac{- 1}{7})$

if $e_{1}$ and $e_{2}$ are eccentricities of two hyperbolas then $p \in ϕ$

if $e_{1}$ and $e_{2}$ are eccentricities of a hyperbola and its conjugate hyperbola then $p \in {0, \frac{- 2}{63}}$

if $e_{1}$ and $e_{2}$ are eccentricities of an ellipse and hyperbola then $p \in (2, \infty)$

answer is A, B.

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Detailed Solution

(a, b)
$\begin{array}{l} (a) D = {(- 8 P)}^{2} - 4 (P - 2) = 4 (16 P^{2} - P + 2) \Rightarrow D > 0 \\ f (1) < 0 \Rightarrow - 7 p - 1 < 0 \Rightarrow p > - \frac{1}{7} \\ f (0) > 0 \Rightarrow p > 2 \Rightarrow p \in (2, \infty) \end{array}$
$\begin{array}{l} (b) f (1) > 0 \Rightarrow p < - \frac{1}{7} \\ \Rightarrow - \frac{- 8 p}{2} > 1 \Rightarrow p > \frac{1}{4} \Rightarrow p \in ϕ \end{array}$
$(c) \frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = 1 \Rightarrow \frac{{(α + β)}^{2} - 2 α β}{{(α β)}^{2}} = 1$
$\Rightarrow 64 p^{2} - 2 p + 4 = p^{2} + 4 - 4 p \Rightarrow 63 p^{2} = - 2 p$
$\Rightarrow p = 0, - \frac{2}{63}$ , one of the roots of the equation is negative
$\begin{array}{l} (d) f (0) > 0 \Rightarrow p > 2 \\ f (1) > 0 \Rightarrow p < - \frac{1}{7} \end{array}} \Rightarrow p \in ϕ$