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Area of an Ellipse Inscribed in a Square of Side 5 sqrt(2) | JEE Advanced Ellipse

JEE Maths question with a full step-by-step solution.

Question
The ellipse x216+y2b2=1\dfrac{x^{2}}{16} + \dfrac{y^{2}}{b^{2}} = 1 is inscribed in a square whose side has length 525\sqrt2. Find the area of the ellipse.
A4π4\pi
B8π8\pi
C10π10\pi
D12π12\picorrect
Solution
Step 1: Here a2=16a^{2} = 16, so a=4a = 4. Each side of the square touches the ellipse and adjacent sides are perpendicular, so every vertex of the square is a point from which two perpendicular tangents are drawn to the ellipse. Step 2: The locus of the intersection of perpendicular tangents to x2a2+y2b2=1\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1 is the director circle
x2+y2=a2+b2=16+b2x^{2}+y^{2} = a^{2}+b^{2} = 16 + b^{2}
 All four vertices lie on this circle, so it is the circumcircle of the square. Step 3: The side is 525\sqrt2, so the diagonal is
52⋅2=10 ⟹ circumradius=55\sqrt2 \cdot \sqrt2 = 10 \ \Longrightarrow\ \text{circumradius} = 5
 Step 4:
16+b2=25 ⟹ b2=9 ⟹ b=316 + b^{2} = 25 \ \Longrightarrow\ b^{2} = 9 \ \Longrightarrow\ b = 3
 b=3<4=ab = 3 < 4 = a, so the ellipse is a genuine one with major axis along xx. Step 5:
Area=πab=π⋅4⋅3=12π\text{Area} = \pi a b = \pi \cdot 4 \cdot 3 = 12\pi
 Answer: (4).
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