3U Parametric Question (1 Viewer)

[Skeptyks]

Just a question I need help with
(I) Prove that the tangent to the parabola at the point P(2t, t^2) has question y-tx-t^2=0 and hence determine the equations of the two tangents to x^2 = 4y through the point (-2,-3).

(II) Show that the line y=mx-2m^2 touches the parabola x^2 = 8y for all values of m, and hence determine the equations of the tangents to the parabola from the point (1, -3).
What are the co-ordinates of the points of contact of these tangents with the curve.

Thanks heaps in advance.

 

[Aysce]

Since I have nothing better to do, I will answer your questions

(I) x^2 = 4y

Now make y the subject,

y = x^2 / 4

So we must now differentiate it to be able to gain the gradient of the tangent.

y' = 2x / 4 = x/2

Since we have the point P, at P(2t, t^2), x = 2t, so lets sub it in the new differentiated equation to gain the gradient of the tangent.

y' = 2t / 2 = t

So we now have the gradient of the tangent which = t.
Now we have a gradient and a point.

Using the point gradient formula:

y-y1 = m(x-x1)

y-t^2 = t(x-2t)

y-t^2 = tx-2t^2

Therefore, y - tx + t^2 = 0. Check if the equation you gave me is precise, I've checked my working over and this is what I received.

I will try to do the other parts later, I am very sleepy and quite rusty with parametrics at the moment, it's been only a month or two but still :L.

 

[zeebobDD]

For part II, sub in the equation y=mx-2m^2 into x^2=8y,

therefore x^2=8(mx-2m^2), then expand and form it into one equation in the form ax^2+bx+c=0

then do discriminant =0 , since the line will only touch the parabola at one point

then yeah pretty much it

 

[thedemonpk3r]

it doesn't work