Conic Sections

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A
alok kumar singh

Contributor-Level 10

e? = √1-b²/25; e? = √1+b²/16
e? = 1
=> (e? )² = 1
=> (1 - b²/25) (1 + b²/16) = 1
=> 1 + b²/16 - b²/25 - b? / (25x16) = 1
=> (9b²)/ (16.25) - b? / (25.16) = 0
=> b²=9
e? = √1-9/25 = 4/5
e? = √1+9/16 = 5/4
α = 2 (5) (e? ) = 8
β = 2 (4) (e? ) = 10
(α, β) = (8,10)

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A
alok kumar singh

Contributor-Level 10

C? = 2C? S = 2√20-4 = 8

 

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V
Vishal Baghel

Contributor-Level 10

∴ center lies on x + y = 2 and in 1st quadrant center = (a, 2-a) where a > 0 and 2-a > 0 ⇒ 0 < a < 2
∴ circle touches x = 3 and y = 2
⇒ |3-a| = |2 - (2-a)| = radius
⇒ |3-a| = |a| ⇒ a = 3/2
∴ radius = a
⇒ Diameter = 2a = 3.

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V
Vishal Baghel

Contributor-Level 10

Let P = (3t², 6t); N = (3t²,0)
M = (3t², 3t)
Equation of MQ: y = 3t
∴ Q = (3/4 t², 3t)
Equation of NQ
y = ( 3t / (3/4 t² - 3t²) ) (x - 3t²)
y - intercept of NQ = 4t = 4/3 ⇒ t = 1/3
∴ MQ = 9/4 t² = 1/4
PN = 6t = 2

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V
Vishal Baghel

Contributor-Level 10

Ellipse: x²/4 + y²/3 = 1
eccentricity = √ (1 - 3/4) = 1/2
∴ foci = (±1,0)
For hyperbola, given 2a = √2 ⇒ a = 1/√2
∴ hyperbola will be x²/ (1/2) - y²/b² = 1
eccentricity = √ (1 + 2b²)
∴ foci = (±√ ( (1+2b²)/2 ), 0)
∴ Ellipse and hyperbola have same foci
√ ( (1+2b²)/2 ) = 1
⇒ b² = 1/2
∴ Equation of hyperbola: x²/ (1/2) - y²/ (1/2) = 1
⇒ x² - y² = 1/2
Clearly, (√3/2, 1/2) does not lie on it.

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

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