Let P 4 , - 4 and Q 9 , 6 be two points on the parabola, y 2 = 4 x and let X be any point on the arc P O Q of this parabola, where O is the vertex of this parabola, such that the area of ∆ P X Q is maximum. Then this maximum area (in sq. units) is :

Step-by-step solution

Two different approaches we can use here. Approach 1: Let X be t 2 ,   2 t , then Area of Δ P X Q = 1 2 t 2 2 t 1 9 6 1 4 - 4 1 Δ = 1 2 . 10 t 2 - t - 6       . . . i For maxima, differentiating both the sides with respect to t and equating it to zero, we get Δ ′ = 0 ⇒ 5 2 t - 1 = 0 ⇒ t = 1 2 Hence, area of Δ P X Q = 5 1 2 2 - 1 2 - 6 = 5 1 4 - 1 2 - 6 = 5 1 - 2 - 24 4 = 125 4 sq.units (using equation i ) Approach 2: For maximum area tangent to the parabola