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

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First, I differentiate x and y
x'=2t
y'=1/t^2

Then, I get 1/t^2 / 2t
= 1/2t^3
Now what do I do?

\[={1\over 2t^3}\]

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