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# Parametric proof watch

1. Hi,

The question is as follows:

Let P be a point on the curve x=t², y=1/t. If the tangent to the curve at P meets the x-axis and y-axis at A and B respectively, prove that PA=2BP.
Seems simple enough - but I end up with PA² = 4t4 + 1/t² and BP² = t4 + 9/4t², the roots of which do not satisfy the requirement.

Could someone show me how it ought to be done?

Cheers!
2. Let P be a point on the curve x=t², y=1/t. If the tangent to the curve at P meets the x-axis and y-axis at A and B respectively, prove that PA=2BP.

Let P be the point where t=p.
Then at P,
Thus the tangent at P has equation:
Or simplified:
At A, when the tangent meets the x axis (y=0) then
At B, when the tangent meets the y axis (x=0) then
Thus we have the information:

If we switch to a "vector style" approach:

Then and
That is, PA=2BP.

Note I've treated PA and BP as vectors because there isn't a modulus sign. In any case, the above still shows |PA|=2|BP| as the vectors PA and 2BP are exactly the same (thus having the same length).

If you haven't looked at vectors yet then you can think of as increasing the x coordinate by and decreasing the y coordinate by as you move from P to A. Note that since we're working in 2 dimensions the coefficient of k doesn't change.
3. (Original post by Theophilus7)
Hi,

The question is as follows:

Seems simple enough - but I end up with PA² = 4t4 + 1/t² and BP² = t4 + 9/4t², the roots of which do not satisfy the requirement.

Could someone show me how it ought to be done?

Cheers!
im bored so will do it using co-ord geom.
using the co-ords from post 2
A=(3p^2,0)
B=(0,3/2p)
general point on curve has co-ords (p^2,1/p)
PA^2=4p^4+1/p^2=(4p^6+1)/p^2
BP^2=p^4+(1/4p^2)=(4p^6+1)/4p^2
PA=rt(4p^6+1)/p
BP=rt(4p^6+1)/2p
giving PA=2BP
4. Thank you both . :-)

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