Conic Sections

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New answer posted

3 weeks ago

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R
Raj Pandey

Contributor-Level 9

4x – 3y + k2 = 0

2 r = 4 0 5 = 8 r = 4

( x 1 ) 2 + ( y + 2 ) 2 = 1 6

New answer posted

3 weeks ago

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A
alok kumar singh

Contributor-Level 10

h = a ( 1 + t 2 ) 2 . . . . . . . ( i )

k = at        ……… (ii)

From (i) & (ii) 2 h a = 1 + k 2 a 2  

required locus of Q is y2 = 2a (x – a/2)

Equation of directrix x – a/2 = -a/2Þ x = 0

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(12)                                                                                                                                                                                      &

...more

New answer posted

3 weeks ago

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V
Vishal Baghel

Contributor-Level 10

cos 20° cos 40° cos 80°
= (2 sin 20° cos 20° cos 40° cos 80°) / (2 sin 20°)
= (2 sin 40° cos 40° cos 80°) / (2² sin 20°)
= (2 sin 80° cos 80°) / (2³ sin 20°)
= 1/8

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

x = t² - t + 1 . (i)
y = t² + t + 1 . (ii)
(i) + (ii)
x + y = 2 (t² + 1) . (iii)
(i) – (ii)
x - y = -2t . (iv)
from (iii) and (iv)
x² – 2xy + y² – 2x – 2y + 4 = 0
Here H² = ab and Δ≠ 0
This is parabola, so e = 1

New answer posted

3 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Equation of the ellipse
(x−3)²/a² + (y+4)²/b² = 1
a=2
ae=1⇒e=1/2
⇒b²=3
Equation of tangent
y+4=m (x−3)±√4m²+3
⇒mx−y=4+3m±√4m²+3
⇒3m±√4m²+3=0
⇒9m²=4m²+3
⇒5m²=3

New answer posted

3 weeks ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Co-ordinate of Q (b+2, a)
⇒ 1/√2 + 7i/√2 = (b+2+ai)e^ (iπ/4)
= (b+2+ai) (cos (π/4)+isin (π/4)
⇒ b−a+2=−1
b+2+a=7
⇒a=4
b=1
⇒2a+b=9

New answer posted

3 weeks ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Let the equation of circle be
x (x-1/2) + y² + λy = 0
=> x² + y² - x/2 + λy = 0
Radius = √ (1/16 + λ²/4) = 2
=> λ² = 63/4 => (x-1/4)² + (y+λ/2)² = 4
∴ This circle and parabola
y-α = (x-1/4)² touch each other, so
α = -λ/2 + 2 => α-2 = -λ/2 => (α-2)² = λ²/4 = 63/16
(4α–8)² = 63

New answer posted

3 weeks ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side
∴ a = √ (3-1)²+ (6-2)² = √20 and
b/ (a/2) = 4/a => b = 8/√5
Area
ab = 2√5 * 8/√5 = 16

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

x²/a² + y²/b² = 1
(-4/√5)²)/a² + (2/√5)²)/b² = 1
=> 32/a² + 9/b² = 1
=> 32/ (5a²) + 9/b² = 1 . (i)
From (i)
6/b² + 9/b² = 1 => b²=15 & a²=16
a²+b² = 15+16=31

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