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Maximum Product of Focal Distances on the Ellipse x^2 + 16y^2 = 4 | JEE Advanced

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

Question
Let AA be a point on the ellipse x2+16y2=4x^{2} + 16y^{2} = 4, whose foci are S1S_1 and S2S_2. Find the maximum value of AS1⋅AS2AS_1 \cdot AS_2.
Solution
Answer: 4 (± 0.01)
Step 1:
x24+y21/4=1 ⟹ a=2,b=12\frac{x^{2}}{4} + \frac{y^{2}}{1/4} = 1 \ \Longrightarrow\ a = 2, \quad b = \frac12
 Step 2: For every point AA on the ellipse,
AS1+AS2=2a=4AS_1 + AS_2 = 2a = 4
 so the sum of the two lengths is fixed while their product varies. Step 3: By AM-GM on the two positive numbers,
AS1⋅AS2≤AS1+AS22=2 ⟹ AS1⋅AS2≤4\sqrt{AS_1\cdot AS_2} \le \frac{AS_1 + AS_2}{2} = 2 \ \Longrightarrow\ AS_1\cdot AS_2 \le 4
 Step 4: Equality needs AS1=AS2=2AS_1 = AS_2 = 2. At an end of the minor axis, A(0,±12)A\left(0,\pm\tfrac12\right), symmetry about the minor axis gives AS1=AS2AS_1 = AS_2, and with sum 44 each is 22. So the bound is attained at a genuine point of the ellipse. Step 5:
(AS1⋅AS2)max⁡=4=4.00\left(AS_1\cdot AS_2\right)_{\max} = 4 = 4.00
 Answer: 4.004.00.
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