# How do you identity if the equation 4x^2+2y^2=8 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Dec 16, 2016

The reference for Conic Section - General Cartesian form tells you how to determine what conic section it is.

#### Explanation:

The reference for Conic Section - General Cartesian form tells you how to determine what conic section it is, when given the General Cartesian form:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

Here is the given equation in the general form:

$2 {x}^{2} + 4 {y}^{2} - 8 = 0$

Please observe the value of ${B}^{2} - 4 A C = {0}^{2} - 4 \left(2\right) \left(4\right) = - 32$. The reference says that this is an ellipse.

Divide both sides of the original equation by 8:

${x}^{2} / 4 + {y}^{2} / 2 = 1$

Write the denominators as squares:

${x}^{2} / {2}^{2} + {y}^{2} / {\left(\sqrt{2}\right)}^{2} = 1$

Insert zeros within the squares in the numerators:

${\left(x - 0\right)}^{2} / {2}^{2} + {\left(y - 0\right)}^{2} / {\left(\sqrt{2}\right)}^{2} = 1$

This is the standard form for an ellipse, because it is easy to see:

1. The center is (0, 0)
2. When $y = 0 , x = - 2 \mathmr{and} 2$
3. When $x = 0 , y = - \sqrt{2} \mathmr{and} \sqrt{2}$

You may use these four points to graph the equation:

$\left(2 , 0\right) , \left(0 , \sqrt{2}\right) , \left(- 2 , 0\right) , \mathmr{and} \left(0 , - \sqrt{2}\right)$

And then sketch in an ellipse around the center.

Here is a graph of the equation: