Let be a point on the parabola , where . The normal to the parabola at meets the -axis at a point . The area of the triangle , where is the focus of the parabola, is 120. If the slope of the normal and are both positive integers, then the pair is

Question

Let $P$ be a point on the parabola $y^{2} = 4 a x$ , where $a > 0$ . The normal to the parabola at $P$ meets the $x$ -axis at a point $Q$ . The area of the triangle $P F Q$ , 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 [2023]

Smart Achievers · Accepted Answer

(1)

$Let point P (a m^{2}, - 2 a m)$

$Equation of normal at P (a m^{2}, - 2 a m) is$

$y = m x - 2 a m - a m^{3}$

$Area of ∆ P F Q$

$= \frac{1}{2} (a + a m^{2}) \times 2 a m = 120$

$\Rightarrow a^{2} m (1 + m^{2}) = 120$

$Pair (a, m) = (2, 3) satisfies the above equation.$

Solution

1 (2, 3)

2 (1, 3)

3 (2, 4)

4 (3, 4)

Ans.
(1)

$Let point P (a m^{2}, - 2 a m)$

$Equation of normal at P (a m^{2}, - 2 a m) is$

$y = m x - 2 a m - a m^{3}$

$Area of ∆ P F Q$

$= \frac{1}{2} (a + a m^{2}) \times 2 a m = 120$

$\Rightarrow a^{2} m (1 + m^{2}) = 120$

$Pair (a, m) = (2, 3) satisfies the above equation.$

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