Question

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 

• A. (2,3)
• B. (1,3)
• C. (2,4)
• D. (3,4)

Answer 1

The Correct Option is A

Equation of normal at P(am², –2am) is y = mx – 2am – am³ 
⇒ Area of ∆PFQ = 1/2(a + am²) x 2am = 120 
a²m(1 + m²) = 120 Pair (a, m) ≡ (2, 3) satisfies above equation.

Answer 2

Given :
y² = 4ax
Now, the equation of the normal is :
y = mx - 2am - am³
Point of Contact are as follows :
P (am², -2am)
Q (2a + am², 0)
So, the area of the △PFQ :
\(=\frac{1}{2}\times|a+am^2||-2am|\)
120 = a²(1 + m²)m
so, a = 2, m = 3
So, only the option (A) satisfies the equation.
So, the correct option is (A) : (2, 3).