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Parametric proof

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!

Reply 1

Gaz031

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.

\frac{dx}{dt}=2t, \frac{dy}{dt}=-\frac{1}{t^2} \rightarrow \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=-\frac{1}{t^2}.\frac{1}{2t}=-\frac{1}{2t^3}

Let P be the point

(p^2,\frac{1}{p})

where t=p.
Then at P,

x=p^2, y=\frac{1}{p}, \frac{dy}{dx}=-\frac{1}{2p^3}

Thus the tangent at P has equation:

y-\frac{1}{p}=-\frac{1}{2p^3}(x-p^2)

Or simplified:

2p^3 y + x = 3p^2

At A, when the tangent meets the x axis (y=0) then

x=3p^2

At B, when the tangent meets the y axis (x=0) then

y=\frac{3}{2p}

Thus we have the information:

A=(3p^2,0)

B=(0,\frac{3}{2p})

P=(p^2, \frac{1}{p})

If we switch to a "vector style" approach:

PA=OA-OP=i(3p^2-p^2)+j(0-\frac{1}{p})+0k=i(2p^2)+j(-\frac{1}{p})+0k

BP=OP-OB=i(p^2-0)+j(\frac{1}{p}-\frac{3}{2p})+0k=i(p^2)+j(-\frac{1}{2p})+0k

Then

PA=i(2p^2)+j(-\frac{1}{p})+0k

and

2BP=i(2p^2)+j(-\frac{1}{p})+0k

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

PA=i(2p^2)+j(-\frac{1}{p})+0k

as increasing the x coordinate by

2p^2

and decreasing the y coordinate by

\frac{1}{p}

as you move from P to A. Note that since we're working in 2 dimensions the coefficient of k doesn't change.

Reply 2

evariste_3086

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

Reply 3

Theophilus7

Thank you both . :-)

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