Q.217. Property of an ellipse.

Question 217.
P is any point on an ellipse, and F is a focus. Show that the circle on PF as diameter touches the auxiliary circle.

Answer 217.
Let P (acosθ, bsinθ) be any point on the ellipse b^2x^2 + a^2y^2 = a^2b^2.
F (ae, 0) is the focus.
Let M be the foot of perpendicular from P on the directrix,
x = a/e.
=> M = (a/e, bsinθ)

=> PF
= ePM
= e(a/e - acosθ)
= a(1 - ecosθ)
=> radius of the circle with PF as diameter,
r1 = PF/2 = a/2 (1 - ecosθ) and center
C1 = [(a/2) (e + cosθ), (b/2)sinθ]

The radius and center of the auxiliary circle are
r2 = a and C2 = (0, 0)

For these two circles to touch each other, r2 - r1 = C1C2. If this condition is proved, then the circles touch each other.

Now, r2 - r1
= a - (a/2) (1 - ecosθ)
= (a/2) (1 + ecosθ) ... (1)

C1C2
= (1/2) √[a^2 (e + cosθ)^2 + b^2 sin^2 θ]
= (1/2) √ (a^2e^2 + 2a^2 ecosθ + a^2 cos^2 θ + b^2 sin^2 θ)
Plugging b^2 = a^2 (1 - e^2),
= (1/2) √ (a^2e^2 + 2a^2 ecosθ + a^2 cos^2 θ + a^2 (1 - e^2) sin^2 θ)
= (1/2) √ (a^2e^2 + 2a^2 ecosθ + a^2 - a^2e^2 sin^2 θ)
= (1/2) √ (a^2e^2 cos^2 θ + 2a^2 ecosθ + a^2)
= (a/2) (1 + ecosθ) ... (2)

From (1) and (2),
r2 - r1 = C1C2
=> the circle with PF as diameter touch the auxiliary circle.

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