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

y^2=8x

cut the

x

-axis at

A

. Let

B(\alpha,\beta)

\alpha>1

, be on the parabola such that the slope of

AB

\tfrac{3}{5}

. If

BC

is a focal chord of the parabola, then six times the area of

\triangle ABC

is:

80

160

correct

174

192

Solution

Step 1: Identify

A

and parametrize

B

For

y^2=8x

a=2

. Directrix:

x=-2

, so

A=(-2,0)

. Let

B=(2t^2,4t)

on the parabola (standard parametrization). Step 2: Use the slope condition to determine

t

m_{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

\alpha = 2t^2 > 1

t=3

gives

B=(18,12)

. (The value

t=\tfrac{1}{3}

gives

\alpha=\tfrac{2}{9}<1

, rejected.) Step 3: Determine the other end

C

of the focal chord For a focal chord,

t_1t_2=-1

, so

t_2=-\tfrac{1}{3}

C = \left(\frac{2}{9},\,-\frac{4}{3}\right)

Step 4: Compute the area of

\triangle ABC

and the required value

\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|

= \frac{1}{2}\left|(-2)\!\left(12+\frac{4}{3}\right)+18\!\left(-\frac{4}{3}\right)+\frac{2}{9}(-12)\right|

= \frac{1}{2}\left|-\frac{80}{3}-24-\frac{8}{3}\right| = \frac{1}{2}\cdot\frac{160}{3} = \frac{80}{3}

6\times\text{Area} = 6\times\frac{80}{3} = 160

Answer: (2)

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