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Ellipse Inscribed in a Square via Director Circle | JEE

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

Question
If the ellipse x2a2−3+y2a+4=1\dfrac{x^2}{a^2-3}+\dfrac{y^2}{a+4}=1 is inscribed in a square of side length a2a\sqrt2 units, then aa is:
A44
B22
C11
Dnone of thesecorrect
Solution
Step 1: The sides of the square are perpendicular tangents to the ellipse, so the square's vertices lie on the director circle. The director circle has radius (a2−3)+(a+4)=a2+a+1\sqrt{(a^2-3)+(a+4)}=\sqrt{a^2+a+1}. Step 2: The square of side a2a\sqrt2 has diagonal a2⋅2=2aa\sqrt2\cdot\sqrt2=2a, which equals the director-circle diameter:
2a2+a+1=2a  ⇒  a2+a+1=a2  ⇒  a=−1.2\sqrt{a^2+a+1}=2a \;\Rightarrow\; a^2+a+1=a^2 \;\Rightarrow\; a=-1.
 Step 3: But for a real ellipse we need a2−3>0a^2-3>0 and a+4>0a+4>0; a=−1a=-1 fails a2−3>0a^2-3>0. So no admissible value exists. Correct answer: (4)
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