In which quadrant does the graph of the parabola $x^2 + y^2 - 2xy - 8x - 8y +32 = 0 $ lie?

One easy way to solve this question would be to simply graph the parabola and check but that;s time consuming when you are provided with around 5 minutes, a pen and a paper and no desmos.

The parabola doesn't intersect the axes. This can be proven by substituting $x=0$ and $y=0$ and confirming that the equation has no real roots.

It can also be represented as $(x-4)^2+ (y-4)^2 = 2xy$. But I am unable to find the quadrant of the parabola through this equation.

I guess if we find the vertex of the parabola, then it's quadrant will be the quadrant of the parabola but how do I find the vertex?

What makes this question difficult is that there are both $x^2$ and $y^2$ present together.

Edit: I see symmetry arguments in the answers, what if it was not so easy to identify the line of symmetry of the parabola? Which method should I employ then?


The question makes it clear that we haven't to check that the curve, let us name it (P), is a parabola.

Here is something that can be done in a short time:

3 successive facts:

  • exchanging $x$ and $y$ doesn't change the equation ; thus parabola (P) is symmetrical with respect to the $y=x$ straight line, thus is its axis.

  • point $S=(2,2)$ belongs to (P).

  • (P) does not intersect the $x$ axis, because, if we set $y=0$ in the equation of (P), we get the quadratic equation $x^2 - 8x +32 = 0$ that has no real roots. Thus, by symmetry, (P) does not intersect the $y$ axis.

Conclusion : (P) lies entirely in the first quadrant.

Remark: As point S is situated on the axis of symmetry of (P), it qualifies it as the apex of (P).


From $x^2+y^2-2xy$ you should directly see that you should let $u=\dfrac1{\sqrt2}(x-y)$ (the $\sqrt2$ is for normalization) and $v=\dfrac1{\sqrt2}(x+y)$.

Then you obtain $2u^2 -8\sqrt2v+32=0$, which becomes $v=\dfrac{\sqrt2}8u^2+2\sqrt2$.

  • $\begingroup$ I upvote your question because you don't deserve downvotes. But you should read more attentively the text: the asker makes it clear that it is assumed that the curve is a parabola and the question is "show that it lies in the first quadrant". $\endgroup$ – Jean Marie Oct 20 '17 at 7:47

Set $u=x+y$ and $v=x-y$. Then $v^2-8u+32=0$ so $v^2= 8(u-4)$. Thus in coordinate system made by $u,v$ focus is at $G(6,0)$ and vertex at $V(4,0)$. The matrix which takes $(x,y)$ to $(u,v)$ is

$$M=\left(% \begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{array}% \right)$$

So $$M^{-1}= {1\over 2}\left(% \begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{array}% \right)$$

So $G$ goes to $F(3,3)$ and vertex $V$ to $O(2,2)$. And since equations $x^2 + 8x +32 = 0 $ and $y^2- 8y +32 = 0 $ don't have the solution, a parabola does not intersect the coordinate lines and so it lies entirely in 1. quadrant.

Hope it is OK because my linear algebra is a little rusty. :(


Well, if you rearrange the equation $x^2 + y^2 - 2xy - 8x - 8y +32 = 0 $, you will get $$(x-y)^2=4\cdot 2\cdot (x+y-4)$$ which is clearly of the form $Y^2=4aX$, where $Y=x-y \,$ and $X=x+y-4$.

So, this is a parabola with axis $y=x$ and vertex at $(2,2)$.

And $x+y=4$ is the tangent to the parabola at the vertex.

Hope, this helps.

  • $\begingroup$ No. $x+y-4=0$ is the tangent at vertex and not the axis. Plot it and check. $y=x$ is the axis. $\endgroup$ – user400242 Oct 17 '17 at 18:53
  • 1
    $\begingroup$ @Blue Thanks for the help. I had just reversed the things. Edited the answer. Again, Thanks!! $\endgroup$ – SchrodingersCat Oct 17 '17 at 18:54
  • $\begingroup$ How would I know before hand that I have to complete the square as $(x-y)^2$ and not $(x-y+1)^2$ or something like that? $\endgroup$ – Archer Oct 17 '17 at 18:59
  • 1
    $\begingroup$ @Abcd 1. Trial and error; 2. Rigorous Practice. $\endgroup$ – SchrodingersCat Oct 17 '17 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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