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
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
(2,3)
(1,3)
(2,4)
(3,4)
The Correct Option is A
Approach Solution - 1
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.
Approach Solution -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).
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When a plane intersects a cone in multiple sections, several types of curves are obtained. These curves can be a circle, an ellipse, a parabola, and a hyperbola. When a plane cuts the cone other than the vertex then the following situations may occur:
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- When α < β < 90°, then the section is an ellipse
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