Asked by sonu
if the line y=1.732(x-1.732) cuts the parabola y^2=x+2 at the point A and B. find PA.PB, where P is the point where line cuts the x-axis.
Answers
Answered by
Reiny
looks like your straight line equation was
y = √3(x - √3) or y = √3x - 3
intersect that with y^2 = x+2
from the straight line: x = (y+3)/√3
y^2 = (y+3)/√3 + 2
√3y^2 = y + 3 + 2√3
√3y^2 - y - (3+2√3) = 0
y = (1 ± √(1 - 4√3(-3-2√3 )/(2√3)
= 2.242 or -1.665 (you had 3 decimal places)
At P, y = 0
x = 1.732 (looks like √3 exactly)
AP = √(2.242-0)^2 + (3.026-1.732)^2 ) = 2.589
PB = √(-1.665-0)^2 + (1.732-.771)^2 ) = 1.922
PA.PB = 4.976
You better check my arithmetic on that one.
y = √3(x - √3) or y = √3x - 3
intersect that with y^2 = x+2
from the straight line: x = (y+3)/√3
y^2 = (y+3)/√3 + 2
√3y^2 = y + 3 + 2√3
√3y^2 - y - (3+2√3) = 0
y = (1 ± √(1 - 4√3(-3-2√3 )/(2√3)
= 2.242 or -1.665 (you had 3 decimal places)
At P, y = 0
x = 1.732 (looks like √3 exactly)
AP = √(2.242-0)^2 + (3.026-1.732)^2 ) = 2.589
PB = √(-1.665-0)^2 + (1.732-.771)^2 ) = 1.922
PA.PB = 4.976
You better check my arithmetic on that one.
Answered by
Prabhakar
It was good question.
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