Parametrics (1 Viewer)

nrlwinner · Feb 11, 2010

$x=t-\frac{4}{t}$
$y=\frac{t^2}{2}+\frac{8}{t^2}$

How can I solve that to get to

$x^2=2(y-4)$

shaon0 · Feb 11, 2010

nrlwinner said:
$x=t-\frac{4}{t}$
$y=\frac{t^2}{2}+\frac{8}{t^2}$

How can I solve that to get to

$x^2=2(y-4)$

x=(t-(4/t))
x^2=(t-(4/t))^2
x^2=(t^2+(16/t^2))-8
x^2+8=t^2+16/t^2
(x^2+8)/2=(t^2)/2+8/t^2
(x^2+8)/2=y
x^2+8=2y
x^2=2(y-4)

nrlwinner · Feb 12, 2010

Got another Q here.

$P(2at,at^2)$ and $Q(2as,as^2)$ are two points on the parabola $x^2=4ay$ . The chord PQ subtends a right angle at the origin. If R is the fourth vertex of the rectangle POQR show that

R parametrically is $x=\frac{2at^2-4}{t},y=\frac{a(t^4+16)}{t^2}$ when st=-4

nrlwinner · Feb 13, 2010

Anyone?

elv09 · Feb 18, 2010

http://www.easy-share.com/1909317349/IMG_0006.jpg

i could be wrong for the x-coordinate ...

Parametrics (1 Viewer)

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