# Trouble finding curve intersection

1. Mar 12, 2008

### silicon_hobo

[SOLVED] trouble finding curve intersection

1. The problem statement, all variables and given/known data
Sketch the area of the region bounded by the curves $$y^2+4x=0$$ and $$y=2x+4$$. Set up two integrals, one with respect to x and one with respect to y, for finding the area of the region. Evaluate one of the integrals to find the area.

2. Relevant equations

3. The attempt at a solution
First I rearrange the equations of the curves to isolate both x & y for each.
$$y^2+4x=0 \Longrightarrow y^2=-4x \Longrightarrow \sqrt{y^2}=\sqrt{-4x} \Longrightarrow y=2\sqrt{-x},-2\sqrt{-x}$$
$$y^2+4x=0 \Longrightarrow x=-\frac{1}{4}y^2$$

$$y=2x+4$$
$$y=2x+4 \Longrightarrow x=\frac{y-4}{2} \Longrightarrow x=\frac{1}{2}y-2$$

Next I must find the domain of the integrals by setting y=y.
$$-\frac{1}{4}y^2=\frac{1}{2}y-2 \Longrightarrow \frac{1}{4}y^2+\frac{1}{2}y-2 \Longrightarrow \frac{1}{2}(y-2)(y+4) \Longrightarrow y=2,-4$$

This is where I'm running into trouble, when I set x=x.
$$2\sqrt{-x}=2x+4$$ $$-2\sqrt{-x}=2x+4$$
I know the points of intersection are -1 & -4 respectively but I can't seem to tease those out of the equations.

$$2\sqrt{-x}=2x+4 \Longrightarrow 2x^\frac{1}{2}+2x+4 \Longrightarrow 2(x^\frac{1}{2}+x+2)$$
$$-2\sqrt{-x}=2x+4 \Longrightarrow -2x^\frac{1}{2}+2x+4 \Longrightarrow 2(-x^\frac{1}{2}+x+2)$$
I'm not sure what to do from here. I can't factor this and if I square it and apply the quadratic equation I get a negative under the sqrt. Some advice would be greatly appreciated. Cheers.

Last edited: Mar 12, 2008
2. Mar 12, 2008

### ircdan

you have
y^2 + 4x = 0
y = 2x + 4

now, y^2 + 4x = 0 => y^2 = -4x => -2x = (y^2)/2

also, y = 2x + 4 => -2x = 4 - y.

So 4 - y = y^2/2, so y^2 + 2y - 8 = 0, so y = - 4 or y = 2, plugging these in
gives that your points of intersection are (-4, -4), (-1, 2).

now just draw the picture and write down your integrals.

3. Mar 12, 2008

Thanks.