Find the Cartesian equation of the curve

[chwala]

Homework Statement

This is a past paper question...see attached (my interest is on the highlighted) I used a different approach thus your insight would be great. I always make it a point to try out the questions before checking what the mark scheme offers...

Relevant Equations

parametric equations.

Find ms solution;

My approach;

$xt=t^2+2$ and $yt=t^2-2$

$xt-2=t^2$ and $yt+2=t^2$

$⇒xt-2=yt+2$

$xt-yt=4$

$t(x-y)=4$

$t=\dfrac{4}{x-y}$

We know that;

$x+y=2t$

$x+y=2⋅\dfrac{4}{x-y}$

$(x-y)(x+y)=8$

$x^2-y^2=8$

Your insight is welcome...rather asking if this approach would be correct.

 

[anuttarasammyak]

x+y=2t
x-y=\frac{4}{t}
(x+y)(x-y)=8

 

[pasmith]

Writing <br /> \begin{split}<br /> x &amp;= t + \frac{2}{t} = 2\sqrt{2}\operatorname{sgn}(t)\cosh(\ln(|t|/\sqrt{2})) \\<br /> y &amp;= t - \frac{2}{t} = 2\sqrt{2}\operatorname{sgn}(t)\sinh(\ln(|t|/\sqrt{2}))\end{split} shows that the curve is a hyperbola, and basic identities then give x^2 - y^2 = (2\sqrt{2})^2 = 8.