Parametrics tangent problem. (1 Viewer)

[shaon0]

P(at^2,2at) is an arbitrary point on the parabola y^2=4ax with Focus S(a,0).
i) Show that the tangent at P has equation x=ty-at^2.

y^2=4ax
d/dx (y^2) = d/dx (4ax)
2y.y' = 4a where y=2at.
y' = 4a/2at.
y' = 1/t but the gradient of the tangent is meant to be t.

I think the question is wrong. Why do i get two different answers?

 

[cyl123]

I think you got confused by the format of the equation you're meant to show. If you rearrange the equation x = ty - at^2, you get:

y = (1/t)x + at which is the gradient intercept form you use to read gradient of.

so the gradient (dy/dx) is meant to be 1/t so theres nothing wrong with the question.

 

[acevipa]

The question seems wrong to me. How can you have a parabola y² = 4ax, yet a focus (0,a)

 

[shaon0]

The question seems wrong to me. How can you have a parabola y² = 4ax, yet a focus (0,a)

sorry that was my bad typing.

 

[Timothy.Siu]

nup, the question is obviously correct, i jsut did it and it was fine and its obvious from inspection

 

[shaon0]

nup, the question is obviously correct, i jsut did it and it was fine and its obvious from inspection

Okay, nw.

 

[addikaye03]

P(at^2,2at) is an arbitrary point on the parabola y^2=4ax with Focus S(a,0).
i) Show that the tangent at P has equation x=ty-at^2.

y^2=4ax
d/dx (y^2) = d/dx (4ax)
2y.y' = 4a where y=2at.
y' = 4a/2at.
y' = 1/t but the gradient of the tangent is meant to be t.

I think the question is wrong. Why do i get two different answers?

well y^2=4ax therefore x=y^2/4a
dx/dy=2y/4a=y/2a; at the arbitary pt (at^2,2at)
dx/dy=2at/2a=t

x-x1=m(y-y1)
x-at^2=t(y-2at)
x-at^2=ty-2at^2
x=ty-at^2#