# Eliminate the parameter and sketch the curve

The question is (in parametric equations):

$$x = 2\sin(t)$$ $$y = \cos(t)$$

for $0 \le t \le \pi/2$

I need to eliminate the parameter and the sketch the curve... any ideas?

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Do you know an ellipse? – Sigur Aug 28 '12 at 2:35

## 1 Answer

Hint: The parametric equation (almost) looks like that of a circle and in particular we have $$\frac{x^2}{2^2} + y^2 = 1$$ Do you know what kind of curve this describes? How much of the curve is traversed? In what direction?

For further reference, see here.

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It looks like an ellipse or circle? – Nick Aug 28 '12 at 2:43
Technically it is an ellipse. Are you familiar with the Cartesian expression for the ellipse? I've added a link that should help a bit. – EuYu Aug 28 '12 at 2:50
Um is it 1/x^2 + 1/y^2 = r? – Nick Aug 28 '12 at 3:15
No! That would give you a degree four curve, namely $x^2 + y^2 - rx^2y^2 = 0.$ An ellipse is a conic section, so has a degree two equation: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. All of the $a$, $b$, $c$, $f$, $g$ and $h$ are fixed numbers of your choice. If $h^2 - ab < 0$ then you have an ellipse, if $h^2-ab = 0$ then you have a parabola. If $h^2 - ab > 0$ then you have a hyperbola. – Fly by Night Aug 28 '12 at 16:45