Straight Lines

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New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

New question posted

a month ago

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Circle x²+y²-2x-4y+4=0
⇒ (x-1)²+ (y-2)²=1
Centre: (1,2) radius=1
line 3x+4y-k=0 intersects the circle at two distinct points.
⇒ distance of centre from the line < radius
⇒ |3*1+4*2-k|/√ (3²+4²) < 1
⇒ |11-k|<5
⇒ 6⇒ k∈ {7,8,9, .,15} since k∈I
Number of K is 9.

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Let L be the common normal to parabola
y = x²+7x+2 and line y = 3x-3
⇒ slope of tangent of y=x²+7x+2 at P=3
⇒ dy/dx|for p = 3
⇒ 2x+7=3 ⇒ x=-2 ⇒ y=-8
So P (-2, -8)
Normal at P: x+3y+C=0
⇒ C=26 (satisfies the line)
Normal: x+3y+26=0

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

(3!)³ (4!)

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.

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