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

JEE Mathematics 2024 Question with Solution

If the points of intersection of two distinct conics x2+y2=4bx^2 + y^2 = 4b and x216+y2b2=1\frac{x^2}{16} + \frac{y^2}{b^2} = 1 lie on the curve y2=3x2y^2 = 3x^2, then 333\sqrt{3} times the area of the rectangle formed by the intersection points is:

  • A

    216216

  • B

    432432

  • C

    108108

  • D

    324324

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The conics are x2+y2=4bx^2 + y^2 = 4b and x216+y2b2=1\frac{x^2}{16} + \frac{y^2}{b^2} = 1. Their intersection points lie on y2=3x2y^2 = 3x^2.

Find: 333\sqrt{3} times the area of the rectangle formed by the intersection points.

Since the intersection points lie on y2=3x2y^2 = 3x^2, substitute this into the first conic:

x2+3x2=4bx^2 + 3x^2 = 4b

So,

4x2=4b⇒x2=b4x^2 = 4b \Rightarrow x^2 = b

Hence,

y2=3by^2 = 3b

Now substitute x2=bx^2 = b and y2=3by^2 = 3b into the second conic:

b16+3bb2=1\frac{b}{16} + \frac{3b}{b^2} = 1

that is,

b16+3b=1\frac{b}{16} + \frac{3}{b} = 1

Multiplying by 16b16b,

b2+48=16bb^2 + 48 = 16b

So,

b2−16b+48=0b^2 - 16b + 48 = 0 (b−4)(b−12)=0(b-4)(b-12) = 0

Thus,

b=4 or 12b = 4 \text{ or } 12

Since the conics are distinct, reject b=4b = 4 because then

x216+y216=1\frac{x^2}{16} + \frac{y^2}{16} = 1

which becomes

x2+y2=16x^2 + y^2 = 16

This is the same as the first conic for b=4b = 4. Therefore,

b=12b = 12

The intersection points are

(±12,±6)(\pm \sqrt{12}, \pm 6)

So the rectangle has side lengths

212and122\sqrt{12} \quad \text{and} \quad 12

Hence its area is

212×12=2412=4832\sqrt{12} \times 12 = 24\sqrt{12} = 48\sqrt{3}

Now compute:

33×483=3×48×3=4323\sqrt{3} \times 48\sqrt{3} = 3 \times 48 \times 3 = 432

Therefore, 333\sqrt{3} times the area of the rectangle is 432432. The correct option is B.

Using symmetry of the intersection points

Given: The common points of the two conics lie on y2=3x2y^2 = 3x^2.

Find: The required scaled area.

Because the equations involve only x2x^2 and y2y^2, every point (x,y)(x,y) gives the symmetric points (±x,±y)(\pm x, \pm y). Thus the four intersection points form a rectangle centered at the origin.

From y2=3x2y^2 = 3x^2 and x2+y2=4bx^2 + y^2 = 4b,

x2+3x2=4bx^2 + 3x^2 = 4b x2=b,y2=3bx^2 = b, \qquad y^2 = 3b

Substitute into x216+y2b2=1\frac{x^2}{16} + \frac{y^2}{b^2} = 1:

b16+3bb2=1\frac{b}{16} + \frac{3b}{b^2} = 1 b16+3b=1\frac{b}{16} + \frac{3}{b} = 1

This gives

b2−16b+48=0b^2 - 16b + 48 = 0

and hence

b=4,12b = 4, 12

Reject b=4b = 4 because the two conics are then not distinct. So b=12b = 12.

Thus the vertices are

(±12,±36)=(±12,±6)(\pm \sqrt{12}, \pm \sqrt{36}) = (\pm \sqrt{12}, \pm 6)

The horizontal side is 2122\sqrt{12} and the vertical side is 1212. Therefore,

Area of rectangle=212⋅12=483\text{Area of rectangle} = 2\sqrt{12} \cdot 12 = 48\sqrt{3}

Hence,

33×Area=33⋅483=4323\sqrt{3} \times \text{Area} = 3\sqrt{3} \cdot 48\sqrt{3} = 432

So the correct option is B.

Common mistakes

  • Taking b=4b = 4 as valid. This is wrong because for b=4b = 4 the second conic becomes the same as the first one, so the conics are not distinct. Always use the condition distinct conics to reject this root.

  • Using the rectangle area incorrectly as the product of coordinates 12×6\sqrt{12} \times 6. This is wrong because those are half-lengths from the origin. Use full side lengths 2122\sqrt{12} and 1212 instead.

  • Substituting y2=3x2y^2 = 3x^2 into only one conic and stopping there. This is wrong because the value of bb must satisfy both conics simultaneously. After finding x2=bx^2 = b, substitute into the second conic as well.

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