Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Find the points of intersection between these two functions.

I haven't done these problems before but I tried to use substitution (treating it like two linear functions, which I have done before).

$$x=2+2y\text{ , } y^2-x^2=2x+4$$ Substituting $x$ into the quadratic function gives: $$y^2-(2+2y)^2=2(2+2y)+4$$ $$-3y^2-12y-12=0$$

I am not sure if this is the correct approach to finding the points of intersection. If so, I don't know the next step. Could someone please explain? Thank you!


Is $-3y^2-12y-12=0$ equivalent to $3y^2+12y+12$? I just moved each term to the other side of the equation...

share|cite|improve this question
up vote 3 down vote accepted

Looks like a good start to me!


$$y^2 + 4y + 4 = 0$$

you can solve for $y$ perhaps by inspection, or by using the quadratic formula:

$$y = \frac{-4 \pm \sqrt{4^2 - 4(1)(4)}}{2} = -2.$$

Then, $x = 2 + 2y = 2 + 2(-2) = -2.$

share|cite|improve this answer
I redid my algebra and noticed I made an error, but I found the point of intersection to be $(-2,-2)$. – Kot Dec 12 '13 at 4:45
Yep, I think that's right. There's also the possibility with other problems that you can have two points of intersection, or no points of intersection, based on whether the part underneath the square root sign is positive or negative, respectively. – John Dec 12 '13 at 4:50
This problem has only one intersection because the part under the root is zero, correct? – Kot Dec 12 '13 at 4:54
That's exactly right. – John Dec 12 '13 at 4:54

Substituting $x$ into the quadratic function gives: \begin{align} y^2 - (2+2y)^2 &= 2(2+2y)+4\\ 3y^2 + 12y + 12 & =0\\ y^2+4y + 4 &= 0 \end{align}

So we obtain $y=-2, x= 2+2y=-2$, $(x,y)=(-2,-2)$, which satisfy the original equations.

From the point of the geometry, the first equation represents a line, and the second represents a hyperbola with center at $(-1,0)$. There is only one point that the two intersect. The point is $(-2,-2)$.enter image description here

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.