Let P be a point on the parabola y² = 4ax, where a > 0 . The normal to the parabola at P meets the x -axis at a point Q. The area of the triangle PFQ , where F is the focus of the parabola, is 120. If the slope m of the normal and $a$ are both positive integers, then the pair $(a, m)$ is

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(2,3)

(1,3)

(2,4)

(3,4)

answer is A.

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Detailed Solution

$P ({at}^{2}, 2 at)$
Equation of normal is tx + y = 2at +at³
$Q (2 a + {at}^{2}, 0) F (a, 0) P ({at}^{2}, 2 at)$
Area of $∆$ PFQ = 120
$\Rightarrow \frac{1}{2} |(a + {at}^{2}) (2 at)| = 120 \Rightarrow |a (1 + t^{2}) (at)| = 120$
Slope of normal = – t
– t, a are integers
$\begin{array}{l} \Rightarrow a^{2} t (1 + t^{2}) = - 120 = - (8) (15) = - 2 (4) (3.5) = - 4 (3) (10) \\ m = - t = 3, a = 2 \end{array}$