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If the tangent at P(at²,2at) to the parabola $y^{2} = 4 a x$ intersects X-axis at ,A and the normal at P meets it at B, then area of triangle PAB is

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$4 a^{2} | t | \sqrt{1 + t^{2}}$

$2 a^{2} | t | (1 + t^{2})$

$4 a^{2} | t | (1 + t^{2})$

$\frac{2 a^{2} (1 + t^{2})}{| t |}$

answer is B.

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Detailed Solution

PA = Length of tangent $= | 2 a t | \sqrt{1 + t^{2}}$

P.B = Length of normal $= | 2 a | \sqrt{1 + t^{2}}$

Area of triangle $P A B = \frac{1}{2} | P A \cdot P B | = 2 a^{2} | t | (1 + t^{2})$

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