Verifying Parametric Equation: (x+y)(x-y)^2 = k

[Hootenanny]

I have a parametric curve defined by the equations;
$$x = t^2 + \frac{2}{t}$$
$$y = t^2 - \frac{2}{t}$$
This is part of a 12 mark question, I have done all the other parts except:

(d) Verify that the Cartesian equation of the curve is $(x+y)(x-y)^2 = k$, stating the value of k.

The previous questions involved finding the co-ordinated at a point, the gradient of the point and hence and equation for the normal at that point.

I know you have to some how isolate t, but I've never done it before when there is two terms.

Help please.

 

[Fermat]

Well, I've just done it, and it's a little bit tricky.

You can simply add the two eqns together, or subtract them, to get expressions for t and/or t².
Then substitute into the original eqns. Thar's when you have to a do little bit of manipulating to work it out.

 

[Hootenanny]

So if I add them I will get
$$t^2 = x + y$$
and
$$t = \sqrt{x+y}$$
Then just sub those in as normal?

 

[Integral]

Start by finding x +y and x -y in terms of t. Where does that take you?

 

[BobG]

So if I add them I will get
$$t^2 = x + y$$
and
$$t = \sqrt{x+y}$$
Then just sub those in as normal?

Check again. I get $$x + y = 2t^2$$

Don't bother solving for t. You want to substitute for 'x+y' and you figured out what to insert into the equation in its place.

Do the same for $$(x-y)^2$$. You'll wind up with an equation whose only variable is t (t^2 actually). Since you're trying to find the constant 'k', you should probably expect 't' to cancel out of your final equation.

It always helps if you have some idea of what type of answer you're looking for.

 

[Hootenanny]

Ive got it now. Thank's a lot!