Conic Sections
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New answer posted
5 months agoContributor-Level 10
Given : a/e = 4 and 1/4 = 1 - b²/a²
Solving : a = 2, b = √3
Parametric co - ordinates are
(2cosθ, √3sinθ) = (1, β)
∴ θ = 60°
∴ Equation of normal is axsecθ − bycosecθ = a² − b²
⇒ 4x - 2y = 1
New answer posted
5 months agoContributor-Level 10
By family of circle, passing through intersection of given circle will be member of S + λS? = 0 family (λ ≠ 1)
(x² + y² – 6x) + λ (x² + y² – 4y) = 0
(λ + 1)x² + (λ + 1)y² – 6x – 4λy = 0
x² + y² - 6/ (λ+1) x - 4λ/ (λ+1) y = 0
Centre (3/ (λ+1), 2λ/ (λ+1)
Centre lies on 2x – 3y + 12 = 0
2 (3/ (λ+1) - 3 (2λ/ (λ+1) + 12 = 0
6λ + 18 = 0
λ = -3
Circle x² + y² – 3x – 6y = 0
New answer posted
5 months agoContributor-Level 10
Since (3,3) lies on x²/a² - y²/b² = 1
9/a² - 9/b² = 1
Now, normal at (3,3) is y-3 = -a²/b² (x-3),
which passes through (9,0) ⇒ b² = 2a²
So, e² = 1 + b²/a² = 3
Also, a² = 9/2
(From (i) and (ii)
Thus, (a², e²) = (9/2, 3)
New answer posted
5 months agoContributor-Level 10
x²/a² + y²/b² = 1 (a > b); 2b²/a = 10 ⇒ b² = 5a
Now, φ (t) = -5/12 + t - t² = 8/12 - (t - 1/2)²
φ (t)max = 8/12 = 2/3 = e ⇒ e² = 1 - b²/a² = 4/9
⇒ a² = 81 (From (i) and (ii)
So, a² + b² = 81 + 45 = 126
New question posted
5 months agoNew answer posted
5 months agoContributor-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)
New answer posted
5 months agoContributor-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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