anonymous
  • anonymous
Let P be a point on the curve x=t^2 , y= 1/t. If the tangent to the curve at P meets the x- and y- axes at A and B respectively, prove that PA=2BP
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
First, I differentiate x and y x'=2t y'=1/t^2
anonymous
  • anonymous
Then, I get 1/t^2 / 2t = 1/2t^3 Now what do I do?
anonymous
  • anonymous
\[={1\over 2t^3}\]

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dumbcow
  • dumbcow
y' = -1/t^2
anonymous
  • anonymous
OH, Yes, I forgot.. so, -1/2t^3
anonymous
  • anonymous
How do I go on from here?
dumbcow
  • dumbcow
well you know the slope of tangent line at P now you need equation of line to find intercepts A and B i think we need to get it in terms of x or unparamaterize it let me play with it a bit
anonymous
  • anonymous
Hm... but there's no value for t, is there?
anonymous
  • anonymous
wouldn't the equation be y= 1/2t^3 +1/2t?
anonymous
  • anonymous
\[y= (1/2t^3)x +1/2t\]
dumbcow
  • dumbcow
ok i think i got it, i was wrong you can leave it terms of t you are on the right track ^^ but the y_intercept is wrong point P is (t^2, 1/t) dy/dx = m = -1/2t^3 tangent line \[y -\frac{1}{t} =- \frac{1}{2t^{3}}(x-t^{2})\] \[y =- \frac{1}{2t^{3}}x +\frac{3}{2t}\] plug in 0 for x and y to get intercepts A and B \[A = (3t^{2},0)\] \[B = (0,\frac{3}{2t})\] to finish proof use distance formula
anonymous
  • anonymous
Ah Ok. Yes, I keep forgetting to add the - sign.
dumbcow
  • dumbcow
@ishaan94 is there another way to do the problem
anonymous
  • anonymous
That's the way it says to do it in my textbook, so I think I should stick with that method :)
dumbcow
  • dumbcow
oh ok
anonymous
  • anonymous
I don't think so, the only method I know is the one you did.

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