If a tangent to the ellipse x² + 4y² = 4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:
If a tangent to the ellipse x² + 4y² = 4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:
Option 1 -
(1, 1)
Option 2 -
(√2, 0)
Option 3 -
(√3, 0)
Option 4 -
(-1, 1)
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1 Answer
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Correct Option - 3
Detailed Solution:Equation of tangent of P (2cosθ, sinθ) is
(cosθ)x + (2sinθ)y = 4
Solving equation of tangent with equation of tangents at major axis ends, i.e. x = -2 and x = 2
For point 'B' (at x=-2):
-2cosθ + 2sinθ y = 4 ⇒ y = (2+cosθ)/sinθ
B (-2, (2+cosθ)/sinθ)
For point 'C' (at x=2):
2cosθ + 2sinθ y = 4 ⇒ y = (2-cosθ)/sinθ
C (2, (2-cosθ)/sinθ)
Now BC is the diameter of circle
Equation of circle: (x+2) (x-2) + (y - (2+cosθ)/sinθ) (y - (2-cosθ)/sinθ) = 0
x²-4 + y² - (4/sinθ)y + (4-cos²θ)/sin²θ...more
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ae = 2b
Or 4 (1 – e2) = e2
4 = 5e2 ->
If two circles intersect at two distinct points
->|r1 – r2| < C1C2 < r1 + r2
| r – 2| < < r + 2
|r – 2| < 5 and r + 2 > 5
–5 < r 2 < 5 r > 3 … (2)
–3 < r < 7 (1)
From (1) and (2)
3 < r < 7
x2 – y2 cosec2q = 5
x2 cosec2q + y2 = 5
and
->
1 + sin2q = 7 – 7 sin2q
->8sin2q = 6
->
->

Slope of axis =
⇒ 2y – 6 = x – 2
⇒ 2y – x – 4 = 0
2x + y – 6 = 0
4x + 2y – 12 = 0
α + 1.6 = 4 ⇒ α = 2.4
β + 2.8 = 6 ⇒ β = 3.2
Ellipse passes through (2.4, 3.2)
⇒
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