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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?

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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).

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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 $v=\dfrac{\sqrt2}8u^2+2\sqrt2$.

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  • $\begingroup$ 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
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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. :(

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @Abcd 1. Trial and error; 2. Rigorous Practice. $\endgroup$ – SchrodingersCat Oct 17 '17 at 19:11

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