$P_{1} : y^{2} = 4 a x, P_{2} : y^{2} = - 4 a x L : y = x$
Area of the rectangle formed by joining the vertices of the latus rectum of the two parabolas $P_{1}$ and $P_{2}$ is

2 a^{2}

sq. units

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4 a^{2}

sq. units

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8 a^{2}

sq. units

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16 a^{2}

sq. units

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Solution

The correct option is C $8 a^{2}$ sq. units
Vertices of latus rectum of $y^{2} = 4 a x$ are $A (a, 2 a)$ and $B (a, - 2 a)$
and Vertices of latus rectum of $y^{2} = - 4 a x$ are $C (- a, 2 a)$ and $D (- a, - 2 a)$
Clearly $A C = B D = 2 a$ and $A B = C D = 4 a$
Area of rectangle $A B C D$ is $= 2 a \times 4 a = 8 a^{2}$ sq. units

Suggest Corrections

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