Logo
Jeeify
MCQMediumJEE 2024Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2024 Question with Solution

Let P be a point on the hyperbola H: x29−y24=1\frac{x^2}{9} - \frac{y^2}{4} = 1, in the first quadrant such that the area of the triangle formed by P and the two foci of H is 13\sqrt{13}. Then, the square of the distance of P from the origin is:

  • A

    1818

  • B

    2626

  • C

    2222

  • D

    2424

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The hyperbola is x29−y24=1\frac{x^2}{9} - \frac{y^2}{4} = 1 and P lies in the first quadrant. The area of the triangle formed by P and the two foci is given in the solution as 2132\sqrt{13}.

Find: The square of the distance of P from the origin.

For the hyperbola x29−y24=1\frac{x^2}{9} - \frac{y^2}{4} = 1, we have

a=3,b=2a = 3, \quad b = 2

so

c=a2+b2=9+4=13c = \sqrt{a^2 + b^2} = \sqrt{9 + 4} = \sqrt{13}

Hence the foci are

(±13,0)\left(\pm \sqrt{13}, 0\right)

Let P=(x,y)P = (x,y). Since PP lies on the hyperbola,

x29−y24=1\frac{x^2}{9} - \frac{y^2}{4} = 1

Using the area formula for the triangle with vertices P(x,y)P(x,y), (13,0)\left(\sqrt{13},0\right) and (−13,0)\left(-\sqrt{13},0\right),

Area=12∣x(0−0)+13(0−y)+(−13)(y−0)∣\text{Area} = \frac{1}{2}\left|x(0-0) + \sqrt{13}(0-y) + (-\sqrt{13})(y-0)\right|

So,

Area=12∣−213y∣=13∣y∣\text{Area} = \frac{1}{2}\left|-2\sqrt{13}y\right| = \sqrt{13}|y|

Now set this equal to the area used in the solution:

13∣y∣=213\sqrt{13}|y| = 2\sqrt{13}

Therefore,

∣y∣=2|y| = 2

Since P is in the first quadrant, y=2y = 2.

Substitute y=2y = 2 into the hyperbola equation:

x29−224=1\frac{x^2}{9} - \frac{2^2}{4} = 1 x29−1=1\frac{x^2}{9} - 1 = 1 x29=2\frac{x^2}{9} = 2 x2=18x^2 = 18

Now the square of the distance of P from the origin is

x2+y2=18+4=22x^2 + y^2 = 18 + 4 = 22

Therefore, the square of the distance of P from the origin is 2222. The correct option is C.

Note: The question states the area as 13\sqrt{13}, but the solution consistently uses 2132\sqrt{13}. The derived answer 2222 follows the solution, which is the authoritative source here.

Common mistakes

  • Using the wrong focal length formula for a hyperbola. For x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, the correct relation is c2=a2+b2c^2 = a^2 + b^2, not a2−b2a^2 - b^2. First find the foci correctly, then use them in the area formula.

  • Forgetting that the area of the triangle with base on the xx-axis depends only on the perpendicular distance of P from the axis. Here the base is the segment joining the foci, so the height is ∣y∣|y|. Do not involve xx unnecessarily in the area after simplification.

  • Ignoring the first quadrant condition. From ∣y∣=2|y| = 2, both y=2y = 2 and y=−2y = -2 are algebraically possible, but only y=2y = 2 satisfies the location of P in the first quadrant.

Practice more Conic Sections (Parabola, Ellipse, Hyperbola) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Conic Sections (Parabola, Ellipse, Hyperbola)

Related questions