Coordinate Systems help!

I was given some questions to practice, my teacher said to go to her for the answers but she's now gone on maternity leave and no teacher can find the answers, I have 3 I'm really struggling with so any help would be appreciated.

1) The general point on the parabola y^2=4ax has coordinates (at^2, 2at). Show that any point on the parabola is equidistant from the point (a,0) and the line x=-a.

Is this just plotting values in?

2) The point P on the rectangular hyperbola xy=c^2 has coordinates (cp, c/p)
i) Find the equation of the normal to the hyperbola at P.
ii) Find the coordinates of the point where the normal meets the hyperbola again.

3) A parabola has Cartesian equation y^2=4ax
i) Find the equation of the normal to the parabola at the point (at^2, 2at)
ii) Verify that the normal at (9a, 6a) passes through the point P(15a, -12a) and find the coordinates of the two other points on the parabola for which the normal passes through P.

Thank you.

Reply 1

ghostwalker

Original post by hannahhh13

For 1.

No; you're looking for an algebraic proof.

A general point on the parabola can be expressed in terms of t; I'm sure you can do that.

Then what's the distance between that point and the point (a,0)? This will be a function of t.

Similarly what's the distance between that point and the line x=-a. If you're unsure, a diagram might help.

Now show they are equal.

Reply 2

hannahhh13

Original post by ghostwalker

Ok, thank you!

Any help with the others? :s-smilie:

Reply 3

hannahhh13

For number 2, I got to y=p^2x-cp^3+c/p :/

Reply 4

Ben121

I got the same as you for number 2, you now have to equate that to the original equation of the hyperbola to get your coords.

Reply 5

hannahhh13

Original post by Ben121

I got the same as you for number 2, you now have to equate that to the original equation of the hyperbola to get your coords.

That's what I'm struggling with, I have xy=c^2 and then y=p^2x-cp^3+c/p but I can't seem to get x or y on it's own

Reply 6

ghostwalker

Original post by hannahhh13

That's what I'm struggling with, I have xy=c^2 and then y=p^2x-cp^3+c/p but I can't seem to get x or y on it's own

Take the value for y from your equation for the normal and substitute it back into the equation for the curve; you'll get a quadratic in x.

It may look horrrible, but remember the values for x which are the roots of this quadratic relate to the points where the normal and the curve intersect, and you know what one of those points is, and hence what one of the roots is; you just need to find the other root.

Reply 7

hannahhh13

Original post by ghostwalker

Thank you, I got x = -c/p^3 and y=-cp^3?

(edited 13 years ago)

Reply 8

ghostwalker

Original post by hannahhh13

Thank you, I got x = -c/p^3 and y=-cp^3?

You can check this yourself by finding if your point satsisfies both the equation for the normal, and the equation for the curve.

If it does, you've found your point, if not, you've made an error.

Have a go at #3 yourself now, it's using techniques you've already applied.