# How do you identity if the equation 4y^2-x^2+4=0 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Jul 10, 2017

I wold say an hyperbola:

Have a look:

Jul 10, 2017

#### Explanation:

When given an equation in the form that you have been given, you can use the general Cartesian form for a conic section:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0 \text{ [1]}$

Please observe that slight rearrangement of the given equation,

$- {x}^{2} + 4 {y}^{2} + 4 = 0 \text{ [2]}$

Fits equation [1] with:

$A = - 1 , B = 0 , C = 4 , D = 0 , E = 0 \mathmr{and} F = 4$

The section of the reference, entitled Discriminant, tells you how to determine what it is:

${B}^{2} - 4 \left(A\right) \left(C\right) = {0}^{2} - 4 \left(- 1\right) \left(4\right)$

${B}^{2} - 4 \left(A\right) \left(C\right) = 16$

$16 > 0$

The fact that the discriminant is greater than 0 tells us that the equation describes a hyperbola.

Subtract for 4 from both sides of equation [2]:

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

Divide both sides by -4:

${x}^{2} / 4 - {y}^{2} = 1 \text{ [3]}$

This fits the standard form ${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [4]}$

${\left(x - 0\right)}^{2} / {2}^{2} - {\left(y - 0\right)}^{2} / {1}^{2} = 1 \text{ [5]}$

Please observe that equation [5] matches the variables of equation [4]:

$h = 0 , k = 0 , a = 2 \mathmr{and} b = 1$

The form for equation [4] is important for graphing because of the following reasons:

1. Everything is centered about the point $\left(h , k\right) = \left(0 , 0\right)$
2. The vertices of the hyperbola are located at $\left(h - a , k\right) = \left(- 2 , 0\right) \mathmr{and} \left(h + a , k\right) = \left(2 , 0\right)$
3. The equations of the asymptotes are $y = - \frac{b}{a} \left(x - h\right) + k \mathmr{and} y = - \frac{b}{a} \left(x - h\right) + k$ which are the two lines $y = - \frac{1}{2} x \mathmr{and} y = \frac{1}{2} x$