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Eccentricity Range: Normal Touching a Circle | JEE

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

Question
Let there exist at least one normal to an ellipse which touches a concentric circle, and the circle intersects the ellipse at four distinct points. Then the eccentricity of the ellipse is:
A(0,12)\left(0,\dfrac{1}{\sqrt2}\right)
B(12,1)\left(\dfrac{1}{\sqrt2},1\right)
C(0,32)\left(0,\dfrac{\sqrt3}{2}\right)
D(32,1)\left(\dfrac{\sqrt3}{2},1\right)correct
Solution
Step 1: For the concentric circle of radius rr to cut the ellipse in four distinct points, b<r<ab<r<a. Step 2: The largest distance from the centre to any normal of the ellipse is a−ba-b. For a normal to touch the circle of radius rr, we need r≤a−br\le a-b. Combined with r>br>b, an admissible rr exists only if
b<a−b  ⇒  a>2b.b<a-b \;\Rightarrow\; a>2b.
 Step 3: Then
e2=1−b2a2>1−b2(2b)2=1−14=34  ⇒  e>32.e^2=1-\frac{b^2}{a^2}>1-\frac{b^2}{(2b)^2}=1-\frac14=\frac34 \;\Rightarrow\; e>\frac{\sqrt3}{2}.
 ∴e∈(32,1)\therefore e\in\left(\dfrac{\sqrt3}{2},1\right). Correct answer: (4)
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