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Focal Chord Length from Inscribed Triangle Area 5a²/2 | JEE

JEE Maths question with a full step-by-step solution.

Question
If the area of the triangle inscribed in the parabola y2=4axy^2 = 4ax with one vertex at the vertex of the parabola and the other two vertices at the extremities of a focal chord is 5a22\dfrac{5a^2}{2}, then the length of the focal chord is:
A3a3a
B5a5a
C25a4\dfrac{25a}{4}correct
DNone of these
Solution
Step 1: Take OO, A(at2,2at)A(at^2, 2at) and the focal partner B(at2,−2at)B\left(\dfrac{a}{t^2}, -\dfrac{2a}{t}\right) (partner of tt is −1t-\tfrac{1}{t}). With a vertex at OO,
Area=12∣xAyB−xByA∣=a2∣t+1t∣.\text{Area} = \frac{1}{2}\left|x_A y_B - x_B y_A\right| = a^2\left|t + \frac{1}{t}\right|.
 Step 2: Set this equal to 5a22\dfrac{5a^2}{2}.
t+1t=52⇒2t2−5t+2=0⇒t=2 or 12.t + \frac{1}{t} = \frac{5}{2} \Rightarrow 2t^2 - 5t + 2 = 0 \Rightarrow t = 2\ \text{or}\ \frac{1}{2}.
 Step 3: The focal chord length is a(t+1t)2a\left(t + \dfrac{1}{t}\right)^2; with t=2t = 2,
a(52)2=25a4.a\left(\frac{5}{2}\right)^2 = \frac{25a}{4}.
 Correct answer: (3)
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