Let the directrix of the parabola P: y^2=8x, cut x-axis at the point A. Let B (alpha, beta), alpha > 1, be a point on P such that the slope of AB is 3/5. If BC is a focal chord of P, then six times the area of triangle ABC is :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q12 – JEE-MAIN 2026

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Home Maths Conic SectionsQuestion #118

Let the directrix of the parabola

P : y^{2} = 8 x

, cut x-axis at the point A. Let B

(α, β)

α > 1

, be a point on P such that the slope of AB is

3 / 5

. If BC is a focal chord of P, then six times the area of

△ A B C

is :

Clue 1 of 3: Understand Parabola Properties

Identify the standard form of the parabola, its focus, and directrix.

🧠 Full Solution Path

Step 1: Identify Parabola Parameters and Point A✦ Active

The given parabola is $y^{2} = 8 x$ . Comparing this with the standard form $y^{2} = 4 a x$ , we find $4 a = 8$ , so $a = 2$ . The directrix of the parabola is $x = - a$ , which means $x = - 2$ . Point A is where the directrix cuts the x-axis, so its coordinates are $A = (- 2, 0)$ . The focus of the parabola is $F = (a, 0) = (2, 0)$ .

Step 2: Determine Coordinates of B and C using Focal Chord Property○ Expand

Let point B be represented by the parametric coordinates $(a t^{2}, 2 a t)$ . Since $a = 2$ , $B = (2 t^{2}, 4 t)$ . We are given $B = (α, β)$ , so $α = 2 t^{2}$ and $β = 4 t$ . The slope of AB is given as $3 / 5$ . Using the coordinates of A and B:

m_{A B} = \frac{4 t - 0}{2 t^{2} - (- 2)} = \frac{4 t}{2 t^{2} + 2} = \frac{2 t}{t^{2} + 1}

Equating this to $3 / 5$ : $\frac{2 t}{t^{2} + 1} = \frac{3}{5} \Rightarrow 10 t = 3 t^{2} + 3 \Rightarrow 3 t^{2} - 10 t + 3 = 0$ . Factoring the quadratic equation, we get $(3 t - 1) (t - 3) = 0$ , which yields $t = 3$ or $t = 1 / 3$ . We are given $α > 1$ , so $2 t^{2} > 1 \Rightarrow t^{2} > 1 / 2$ . If $t = 1 / 3$ , then $t^{2} = 1 / 9$ , which is less than $1 / 2$ , so this value is rejected. Thus, $t = 3$ . The coordinates of B are $B = (2 (3^{2}), 4 (3)) = (18, 12)$ . Since BC is a focal chord, if B corresponds to parameter $t$ , C corresponds to parameter $- 1 / t$ . So, for $t = 3$ , C corresponds to $t^{'} = - 1 / 3$ . The coordinates of C are $C = (2 (- 1 / 3)^{2}, 4 (- 1 / 3)) = (2 / 9, - 4 / 3)$ .

💡 Teacher's Secret Hint

Remember to check all given conditions, like $α > 1$ , to select the correct parameter value.

Step 3: Calculate Area of

△ A B C

and the Final Value○ Expand

The vertices of the triangle are $A = (- 2, 0)$ , $B = (18, 12)$ , and $C = (2 / 9, - 4 / 3)$ . The area of $△ A B C$ can be calculated using the determinant formula:

Area = \frac{1}{2} | x_{A} (y_{B} - y_{C}) + x_{B} (y_{C} - y_{A}) + x_{C} (y_{A} - y_{B}) |

Substituting the coordinates:

Area = \frac{1}{2} | - 2 (12 - (- 4 / 3)) + 18 (- 4 / 3 - 0) + \frac{2}{9} (0 - 12) |

Area = \frac{1}{2} | - 2 (12 + 4 / 3) - 24 - \frac{24}{9} |

Area = \frac{1}{2} | - 2 (\frac{36 + 4}{3}) - 24 - \frac{8}{3} |

Area = \frac{1}{2} | - \frac{80}{3} - \frac{72}{3} - \frac{8}{3} |

Area = \frac{1}{2} | - \frac{160}{3} | = \frac{1}{2} \times \frac{160}{3} = \frac{80}{3}

We need to find six times the area of $△ A B C$ :

6 \times Area = 6 \times \frac{80}{3} = 2 \times 80 = 160

💡 Teacher's Secret Hint

Be careful with arithmetic, especially when dealing with fractions and negative signs in the area calculation.

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Maths Question 12 – JEE-MAIN 2026

Clue 1 of 3: Understand Parabola Properties

🧠 Full Solution Path