ParabolamediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Parabola: Let Directrix Cut Axis Let Parabola Slope Focal (JEE Main 2026)

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

Question
Let the directrix of y2=8xy^2=8x cut the xx-axis at AA. Let B(α,β)B(\alpha,\beta), α>1\alpha>1, be on the parabola such that the slope of ABAB is 35\tfrac{3}{5}. If BCBC is a focal chord of the parabola, then six times the area of △ABC\triangle ABC is:
A8080
B160160correct
C174174
D192192
Solution
Step 1: Identify AA and parametrize BB For y2=8xy^2=8x: a=2a=2. Directrix: x=−2x=-2, so A=(−2,0)A=(-2,0). Let B=(2t2,4t)B=(2t^2,4t) on the parabola (standard parametrization). Step 2: Use the slope condition to determine tt
mAB=4t−02t2+2=2tt2+1=35  ⟹  3t2−10t+3=0  ⟹  (3t−1)(t−3)=0m_{AB} = \frac{4t-0}{2t^2+2} = \frac{2t}{t^2+1} = \frac{3}{5} \implies 3t^2-10t+3=0 \implies (3t-1)(t-3)=0
 Since α=2t2>1\alpha = 2t^2 > 1: t=3t=3 gives B=(18,12)B=(18,12). (The value t=13t=\tfrac{1}{3} gives α=29<1\alpha=\tfrac{2}{9}<1, rejected.) Step 3: Determine the other end CC of the focal chord For a focal chord, t1t2=−1t_1t_2=-1, so t2=−13t_2=-\tfrac{1}{3}:
C=(29, −43)C = \left(\frac{2}{9},\,-\frac{4}{3}\right)
 Step 4: Compute the area of △ABC\triangle ABC and the required value
Area=12∣xA(yB−yC)+xB(yC−yA)+xC(yA−yB)∣\text{Area} = \frac{1}{2}\left|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\right|
=12∣(−2) ⁣(12+43)+18 ⁣(−43)+29(−12)∣= \frac{1}{2}\left|(-2)\!\left(12+\frac{4}{3}\right)+18\!\left(-\frac{4}{3}\right)+\frac{2}{9}(-12)\right|
=12∣−803−24−83∣=12⋅1603=803= \frac{1}{2}\left|-\frac{80}{3}-24-\frac{8}{3}\right| = \frac{1}{2}\cdot\frac{160}{3} = \frac{80}{3}
6×Area=6×803=1606\times\text{Area} = 6\times\frac{80}{3} = 160
 Answer: (2)
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