The Student Room Group

Reply 1

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.

dxdt=2t,dydt=−1t2→dydx=dydt.dtdx=−1t2.12t=−12t3\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 (p2,1p)(p^2,\frac{1}{p}) where t=p.
Then at P,x=p2,y=1p,dydx=−12p3x=p^2, y=\frac{1}{p}, \frac{dy}{dx}=-\frac{1}{2p^3}
Thus the tangent at P has equation: y−1p=−12p3(x−p2)y-\frac{1}{p}=-\frac{1}{2p^3}(x-p^2)
Or simplified:2p3y+x=3p22p^3 y + x = 3p^2
At A, when the tangent meets the x axis (y=0) then x=3p2x=3p^2
At B, when the tangent meets the y axis (x=0) then y=32py=\frac{3}{2p}
Thus we have the information:
A=(3p2,0)A=(3p^2,0)
B=(0,32p)B=(0,\frac{3}{2p})
P=(p2,1p)P=(p^2, \frac{1}{p})
If we switch to a "vector style" approach:
PA=OA−OP=i(3p2−p2)+j(0−1p)+0k=i(2p2)+j(−1p)+0kPA=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(p2−0)+j(1p−32p)+0k=i(p2)+j(−12p)+0kBP=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(2p2)+j(−1p)+0kPA=i(2p^2)+j(-\frac{1}{p})+0k and 2BP=i(2p2)+j(−1p)+0k2BP=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(2p2)+j(−1p)+0kPA=i(2p^2)+j(-\frac{1}{p})+0k as increasing the x coordinate by 2p22p^2 and decreasing the y coordinate by 1p\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

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

Thank you both . :-)