Straight Lines

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New answer posted

4 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

By using condition of tangency,
c² = a² (m² + 1)
⇒ c² = 5 [ (2)² + 1]
⇒ c² = 25
⇒ c = ±5

New answer posted

4 weeks ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 weeks ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

2x + y = 4
2x + 6y = 14
} y=2, x=3
B (1, 2)
Let C (k, 4–2k)
Now AB² = AC²
=> 5k² – 24k + 19 = 0
α = (6+1+10/5)/3 = 18/5
Now 15 (α+β)
15 (17/5) = 51

New answer posted

4 weeks ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

dy/dx = (ax-by+a)/ (bx+cy+a)
=> bxdy + cydy + ady = axdx – bydx + adx
cy²/2 + ay – ax²/2 – ax + bxy = k
ax² + ay² + 2ax – 2ay = k
=> x² + y² + 2x – 2y = λ
Short distance of (11,6)
= √12²+5² – 5
= 13 – 5
= 8

New answer posted

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Let z = αi + βj be a complex number & √3i + j = (√3 + i) where I = √-1.
|z| = |z - (√3 + i)| and |z| = |√3 + i| = 2.
This implies z lies on the perpendicular bisector of the segment from 0 to √3+i and on a circle of radius 2 centered at the origin.
z = (√3 + i) ( (1+i)/√2 )
z = (1/√2) [ (√3 - 1) + I (√3 + 1)]
∴ α = (√3 - 1)/√2, β = (√3 + 1)/√2
Area of required triangle = (1/2) * base * height = (1/2) * |α| * |β| * 2 (This seems to be area of triangle with vertices 0, z, and another point)
The provided calculation: Area = (1/2) * | (√3-1)/√2| * | (√3+1)/√2| = (1/2) * (3-1)/2 = 1/2.

New answer posted

a month ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Parabola: y² = 4x - 20 = 4(x - 5). Vertex at (5,0).
Line: The text seems to derive the tangent equation y = x - 4. This is not a tangent to the given parabola. The standard tangent to y²=4aX is Y=mX+a/m. Here X=x-5, a=1. So y = m(x-5)+1/m.
The other curve is an ellipse: x²/a² + y²/b² = 1.
The text says x²/2 + (x-4)²/b² = 1. This assumes a² = 2.
x²/2 + (x²-8x+16)/b² = 1
x²(1/2 + 1/b²) - (8/b²)x + (16/b² - 1) = 0.
For tangency, the discriminant (D) of this quadratic equation must be zero.
D = (8/b²)² - 4(1/2 + 1/b²)(16/b² - 1) = 0.
64/b? - 4(8/b² - 1/2 + 16/b? - 1/b²) = 0.
16/b? - (7/b² - 1/2 + 16/b?) = 0.
-7/b² + 1/2 = 0

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New question posted

a month ago

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