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Let $P_{1}$ and $P_{2}$ be two parabolas with distinct directrices $l_{1}$ and $l_{2}$ and distinct foci $F_{1}$ and $F_{2}$ respectively. It is known that $F_{1} F_{2}$ is parallel to $l_{1}$ and $l_{1}$ and $l_{1}$ are parallel. $F_{1}$ lies on $P_{2}$ and $F_{2}$ lies on $P_{1}$ . The two parabolas intersect at distinct points A and B.
Given that $F_{1} F_{2} = 1$ , the value of $A B^{2}$ can be

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$\frac{15}{4}$

answer is D.

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Detailed Solution

$P_{1} : y^{2} = 4 a x P_{2} : {(y - k)}^{2} = - 4 b (x - 4) F_{1} (a, 0) F_{2} (h - b, k)$

$F_{1} F_{2} p a r a l l e l t o l_{1}, l_{2} \Rightarrow h - b = a$

$\Rightarrow h = a + b$

$F_{1} l i e s o n P_{2} \Rightarrow k^{2} = - 4 b (a - b) = 4 b^{2}____(1) F_{2} l i e s o n P_{1} \Rightarrow k^{2} = 4 a (h - b) = 4 a^{2}____(2)$

$| a | = | b | F_{1} (a, 0) F_{2} (a, \pm 2 a) \Rightarrow F_{1} F_{2} = 2 a$

A and B are points of intersections of $y^{2} = 4 a x$ and ${(y - 2 a)}^{2} = - 4 a (x - 2 a)$

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