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Arc Length of Centre Locus of a Sliding Ellipse | JEE

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

Question
An ellipse with major and minor axes of length 10310\sqrt3 and 1010 units slides along the coordinate axes and always remains confined in the first quadrant. The locus of the centre of the ellipse is the arc of a circle. The length of this arc is:
A10π10\pi units
B5π5\pi units
C54π\dfrac{5}{4}\pi units
D53π\dfrac{5}{3}\pi unitscorrect
Solution
Step 1: Find aa and bb. The semi-axes are a=1032=53a=\dfrac{10\sqrt3}{2}=5\sqrt3 and b=102=5b=\dfrac{10}{2}=5. Step 2: While both axes stay tangent to the ellipse, the centre lies on the director circle, whose radius from the origin is
R=a2+b2=75+25=10.R=\sqrt{a^2+b^2}=\sqrt{75+25}=10.
 So the centre moves on a circle of radius 1010. Step 3: Find the angle of the arc. Between the two extreme positions,
θ=∠C1OC2=π2−2tan⁡−1 ⁣(553)=π2−2tan⁡−1 ⁣13=π2−2⋅π6=π6.\theta=\angle C_1OC_2=\frac{\pi}{2}-2\tan^{-1}\!\left(\frac{5}{5\sqrt3}\right)=\frac{\pi}{2}-2\tan^{-1}\!\frac{1}{\sqrt3}=\frac{\pi}{2}-2\cdot\frac{\pi}{6}=\frac{\pi}{6}.
 Step 4: The arc length is
Rθ=10⋅π6=5π3 units.R\theta=10\cdot\frac{\pi}{6}=\frac{5\pi}{3}\text{ units}.
 Correct answer: (4)
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