anonymous
  • anonymous
The parabola y=ax^2+bx+c, c≠0 meets the x-axis at A(α,0) and B(β,0) and the y-axis at C. If AC and BC are perpendicular, prove ac=-1
Mathematics
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anonymous
  • anonymous
The parabola y=ax^2+bx+c, c≠0 meets the x-axis at A(α,0) and B(β,0) and the y-axis at C. If AC and BC are perpendicular, prove ac=-1
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
alpha and beta are the roots of quadratic ax^2+bx+c=0
anonymous
  • anonymous
so alpha + beta = -b/a and alpha* beta= c/a

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lgbasallote
  • lgbasallote
sure...tag the math hater to a math problem...makes a lot of sense
anonymous
  • anonymous
slope of AC * slope of BC= -1
anonymous
  • anonymous
so alpha* beta = -c^2
anonymous
  • anonymous
and put this value into... alpha* beta= c/a
anonymous
  • anonymous
so ac=-1
anonymous
  • anonymous
does it make sense?
anonymous
  • anonymous
how did you get to alpha*beta=-c^2?
anonymous
  • anonymous
use this...slope of AC * slope of BC= -1 c=(0, beta)
mathslover
  • mathslover
No one tags mathslover though :P
anonymous
  • anonymous
since when does c=(0,beta)?
anonymous
  • anonymous
so slope of AC= - c/alpha=m1 , slope of BC= -c/beta=m2 m1 *m2= -1...since AC is perpendicular to BC.
anonymous
  • anonymous
sorry C=(0,c)
anonymous
  • anonymous
right
anonymous
  • anonymous
anonymous
  • anonymous
u got it?
anonymous
  • anonymous
yah

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