Start free

Mathematics · Ellipse

JEE Main 2025 — 3 April, Evening Shift — Question 40

Let CC be the circle of minimum area enclosing the ellipse E:x2a2+y2b2=1E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 with eccentricity 12\frac{1}{2} and foci (±2,0)( \pm 2,0). Let PQRP Q R be a variable triangle, whose vertex PP is on the circle CC and the side QRQ R of length 2 a is parallel to the major axis of EE and contains the point of intersection of EE with the negative yy-axis. Then the maximum area of the triangle PQRP Q R is:

  1. Option A:

    8(2+3)8(2+\sqrt{3})

    Correct
  2. Option B:

    6(2+3)6(2+\sqrt{3})

  3. Option C:

    6(3+2)6(3+\sqrt{2})

  4. Option D:

    8(3+2)8(3+\sqrt{2})

Answer: A

Step-by-step solution

Area =12×(a+b)⋅2a=a(a+b)=\frac{1}{2} \times(a+b) \cdot 2 a=a(a+b)

Since, e=12ae=2⇒a=4,b=23e=\frac{1}{2} a e=2 \Rightarrow a=4, b=2 \sqrt{3}

⇒\Rightarrow Area =4(4+23)=8(2+3)=4(4+2 \sqrt{3})=8(2+\sqrt{3})

Solution figure

Answer key and solution verified before publishing.

Practise Ellipse

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
3 April, Evening Shift
Subject
Mathematics
Chapter
Ellipse
Topic
Special properties of ellipse
Let C be the circle of minimum area enclosing the ellipse E: frac x 2… | JEE Main 2025 PYQ with Solution · DhiX AI