# If $x$ and $y$ are two real quantities connected by the equation $9x^2+2xy+y^2-92x-20y+244=0$ then will $x \in [3,6]$, and $y \in [1,10]$?

If $$x$$ and $$y$$ are two real quantities connected by the equation $$9x^2+2xy+y^2-92x-20y+244=0$$ then will $$x \in [3,6]$$ , and $$y \in [1,10]$$ ?
What could be some of the intuitive / standard ways of finding that ?

What I've tried:

My attempt using completing squares :

The given $$eq^n$$ can be expressed as $$8(x-3)(x-6) + (x+y-10)^2 = 0$$

From here , it's clear that $$(x+y-10)^2 ≥ 0$$ and for that $$(x-3)(x-6) ≤ 0$$ which implies $$x \in [3,6]$$ and $$|x+y-10| ≥ 0$$

Now since, $$|x+y-10| ≥ 0$$ and $$x \in [3,6]$$ therefore, $$y \in [4,7]$$ . Where am I being wrong ?

• Form a quadratic in $x$ then do discriminant>0 find the range of y....do same for x Apr 16, 2023 at 12:51
• You could first try to draw a graph with any software you like. Apr 16, 2023 at 13:20
• Plot the equation and find the coordinates of the major and minor axes of the ellipse Apr 16, 2023 at 13:35
• Of course, $|x+y-10|\ge 0$ and $x\in [3,6]$ tell nothing about the range of $y$. $|x+y-10|\ge 0$ is not $|x+y|\le 10$. Apr 16, 2023 at 17:54
• @Ash_Blanc I think dxiv's answer is better than mine (because he simply completed the square in $y$ which is more natural). Please don't accept my answer (Thanks for accepting mine). May 10, 2023 at 15:35

If one is allowed to use the Lagrange multiplier then this is fairly standard.

Let $$F(x,y,\lambda) = x + \lambda(9x^2+2xy+y^2-92x-20y+244)$$, then $$F'_x = 2\lambda(9x+y-46)+1, F'_y = 2\lambda(x+y-10),$$ so letting $$F'_x = F'_y = 0$$ we get $$x = (72\lambda-1)/(16\lambda), y = (88\lambda+1)/(16\lambda)$$, and $$0 = 9x^2+2xy+y^2-92x-20y+244 = (1 - 576\lambda^2)/(32\lambda^2)$$, which implies that $$\lambda = \pm 1/24$$, and this corresponds to $$x=3$$ or $$x=6$$.

Similarly, let $$G(x,y,\lambda) = y + \lambda(9x^2+2xy+y^2-92x-20y+244)$$, then $$G'_x = 2\lambda(9x+y-46), G'_y = 2\lambda(x+y-10)+1,$$ so $$G'_x = G'_y = 0$$ yields $$x = (72\lambda+1)/(16\lambda), y = (88\lambda-9)/(16\lambda)$$, and $$0 = 9x^2+2xy+y^2-92x-20y+244 = (9 - 576\lambda^2)/(32\lambda^2)$$, so $$\lambda = \pm 1/8$$, and this corresponds to $$y=1$$ or $$y=10$$.

• Well @Jianing Song except this one Apr 16, 2023 at 13:18
• @Ash_Blanc So you have ruled out two ways that seem to be most natural for me. I would then suggest parametrizing. Apr 16, 2023 at 13:21
• @Ash_Blanc You should say explicitly what answer you are looking for, instead of wasting someone's time and letting them post an answer that you don't want. Apr 16, 2023 at 13:24
• @Ash_Blanc Write you equation in the form $(\cdots)^2 + (\cdots)^2 = \cdots$, then use trigonometric substitution. I'm leaving that to you :) Apr 16, 2023 at 13:30
• @Ash_Blanc if you're looking for high school level solutions then attach the tag algebra precalculus...assigning limits as a tag suggests that you're familiar with calculus Apr 16, 2023 at 13:34

The given $$eq^n$$ can be expressed as $$8(x-3)(x-6) + (x+y-10)^2 = 0$$

From here , it's clear that $$(x+y-10)^2 ≥ 0$$ and for that $$(x-3)(x-6) ≤ 0$$ which implies $$x \in [3,6]$$

That works. Or, along the same idea, just with a different turn at the end:

\begin{align} 0 &= 9x^2 + 2xy + y^2 - 92x - 20y + 244 \\ &= y^2 + 2(x - 10)y \color{red}{+ (x - 10)^2 - (x - 10)^2} + 9x^2 - 92x + 244 \\ &= (y + x - 10)^2 + 8 x^2 - 72 x + 144 \\ &= (x + y - 10)^2 + 8 \left(x - \frac{9}{2}\right)^2 - 18 \\ \iff &\quad(x + y - 10)^2 + 8 \left(x - \frac{9}{2}\right)^2 = 18 \end{align}

It follows that:

$$8 \left(x - \frac{9}{2}\right)^2 \le 18 \;\iff\; \left|x - \frac{9}{2}\right| \le \sqrt{\frac{18}{8}} = \frac{3}{2} \;\iff\; 3 = \frac{9}{2} - \frac{3}{2}\le x \le \frac{9}{2} + \frac{3}{2} = 6$$

[ EDIT ] At this point, knowing the range of $$x$$, it may be tempting to try and use the same argument for $$|x + y - 10| \le \sqrt{18} = 3\sqrt{2}$$ $$\iff 10 - 3\sqrt{2} - x \le y \le 10 + 3\sqrt{2} - x$$ $$\implies y \in [4 - 3\sqrt{2}, 7 + 3\sqrt{2}]$$. This is not wrong, but the bounds are not tight, either, and the interval is not the true range of $$y$$ because some values can never be attained. That's because the two terms $$(x + y - 10)$$ and $$x - \dfrac{9}{2}$$ are not independent, and the extrema of $$y$$ are not attained at the same points where the extrema of $$x$$ are attained. In fact, as shown below, the actual range of $$y$$ is $$[1,10] \subsetneqq [4 - 3\sqrt{2}, 7 + 3\sqrt{2}]$$.

since, $$|x+y-10| ≥ 0$$ and $$x \in [3,6]$$ therefore, $$y \in [4,7]$$. Where am I being wrong?

The "therefore" step is not justified, and the implication is wrong. The inequality $$|x+y-10| \ge 0$$ holds true no matter what the values of $$x,y$$ are, because an absolute value is always non-negative.

But, remember the idea that worked the first time around: you completed the square in $$y$$, and you found the range of $$x$$. Now, try to complete the other square in $$x$$, and you'll get the range of $$\,y\,$$:

\begin{align} 0 &= 9x^2 + 2xy + y^2 - 92x - 20y + 244 \\ &= (3x)^2 + 2\,\frac{y-46}{3} \, 3x \color{red}{+ \left(\frac{y-46}{3}\right)^2 - \left(\frac{y-46}{3}\right)^2} + y^2 - 20y + 244 \\ &= \left(3x + \frac{y-46}{3}\right)^2 + \frac{8}{9} y^2 - \frac{88}{9}y + \frac{80}{9} \\ &= \left(3x + \frac{y-46}{3}\right)^2 + \frac{8}{9} (y-1)(y-10) \end{align}

It follows that:

$$(y-1)(y-10) \le 0 \;\;\iff\;\; 1 \le y \le 10$$

• Your solution is nice. (+1) May 10, 2023 at 15:31
• Nice. But, I think you need to mention that, completing the square just a hidden version of the Discriminant method. By finding the discriminant, you have already completed the square. No special magic here. Rather, I think completing the square is more work.
– User
May 10, 2023 at 16:14
• @RiverLi Thanks. I thought this was a well asked question, which deserved all the good answers it got.
– dxiv
May 10, 2023 at 16:38
• @User Thanks, and you are right. However, the way I wrote it follows more closely OPs attempt, in order to show what they got right, where things went wrong, and how the idea could be salvaged.
– dxiv
May 10, 2023 at 16:39
– User
May 10, 2023 at 16:44

GEOMETRIC APPROACH

Quadratic equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ represents a conic section, which may be degenerate. The given equation is $$9x^2+2xy+y^2-92x-20y+244=0.$$

With the use of matrix representation, compute the determinant of the associate matrix $$\left|{\begin{matrix}A & B/2 & D/2\\B/2 & C & E/2\\ D/2 & E/2 & F\end{matrix}}\right|=\left|{\begin{matrix}9 & 1 & -46\\1 & 1 & -10\\ -46 & -10 & 244\end{matrix}}\right|=-144\neq0,$$ and $$\left|{\begin{matrix}A & B/2 \\B/2 & C\end{matrix}}\right|=\left|{\begin{matrix}9 & 1 \\1 & 1 \end{matrix}}\right|=8>0,$$

therefore $$9x^2+2xy+y^2-92x-20y+244=0$$ is equation of an ellipse. It is enclosed in a rectangle with horizontal and vertical sides, which lie on horizontal and vertical tangents to this ellipse, respectively.

Horizontal tangents
We can use a simple high-school approach: find a horizontal line with equation $$y=c$$ that has a unique common point with the ellipse. This leads to the quadratic equation $$9x^2+x(2c-92)+(c^2-20c+244)=0,$$ whose discriminant $$\Delta=-16(2c^2-22c+20)$$ must be $$0.$$
The two horizontal tangents have equations $$y=1,\quad y=10.$$ It follows necessarily that $$1\le y\le10$$ for each point on the ellipse, because ellipse is convex.

A similar computation gives two vertical tangents $$x=3, \quad x=6,$$ which proves that $$3\le x\le 6$$ for any point on the ellipse.

Trigonometric substitution should help (as Jianing Song commented).

We can write $$9x^2+2xy+y^2-92x-20y+244=0$$ as $$y^2+2(x-10)y+9x^2-92x+244=0$$ $$(y+x-10)^2-(x-10)^2+9x^2-92x+244=0$$ $$(y+x-10)^2+8x^2-72x+144=0$$ $$(x + y - 10)^2 + 8 \left(x - \frac{9}{2}\right)^2 = 18$$

Now, dividing the both sides by $$18$$, we get $$\frac{1}{18}(x + y - 10)^2 + \frac 49 \left(x - \frac{9}{2}\right)^2 = 1$$ which can be written as $$\bigg(\frac{x+y-10}{3\sqrt 2}\bigg)^2+\bigg(\frac 23x-3\bigg)^2=1$$

Here, let us use trigonometric substitution : $$\cos t=\frac{x+y-10}{3\sqrt 2},\qquad \sin t=\frac 23x-3\tag1$$ Solving $$(1)$$ for $$x,y$$ gives $$x=\frac 32\sin t+\frac 92$$ and $$y=3\sqrt 2\ \cos t-\frac 32\sin t+\frac{11}2=\frac 92\cos(t+\alpha)+\frac{11}{2}$$ where $$\alpha$$ is such that $$\cos\alpha=\frac{2\sqrt 2}{3}$$ and $$\sin\alpha=\frac 13$$.

Therefore, we finally obtain $$3=\frac 32(-1)+\frac 92\leqslant x\leqslant \frac 32\cdot 1+\frac 92=6$$ and $$1=\frac 92(-1)+\frac{11}{2}\leqslant y\leqslant \frac 92\cdot 1+\frac{11}{2}=10$$

Let $$F(x.y)=9x^2+2xy+y^2-92x-20y+244=0,$$ then $$\dfrac{\text dy}{\text dx}=-\dfrac{F'_x}{F'_y}= \dfrac{18x+2y-92}{2x+2y-20}.$$ $$\dfrac{\text dx}{\text dy}=-\dfrac{F'_y}{F'_x}= \dfrac{2x+2y-20}{18x+2y-92}.$$ I.e., extremes by $$y$$ are defined by the system $$\begin{cases} 18x+2y-92=0\\ 9x^2+2xy+y^2-92x-20y+244=0, \end{cases}$$ $$\begin{cases} y=46-9x\\ 9x^2+2x(46-9x)+y(46-9x)^2-92x-20(46-9x)+244=0, \end{cases}$$ $$\begin{cases} y=46-9x\\ 72(x-4)(x-5)=0, \end{cases}$$ $$\dbinom xy\in\left\{\dbinom4{10},\dbinom51\right\},$$ $$\color{brown}{\mathbf{y\in[1,10].}}$$

Extremes by x are defined by the system $$\begin{cases} 2x+2y-20=0\\ 9x^2+2xy+y^2-92x-20y+244=0, \end{cases}$$ $$\begin{cases} x=10-y\\ 9(10-y)^2+2y(10-y)+y^2-92(10-y)-20y+244=0 \end{cases}$$ $$\begin{cases} x=10-y\\ 8(y-4)(y-7)=0, \end{cases}$$ $$\dbinom xy\in\left\{\dbinom64,\dbinom37\right\},$$ $$\color{brown}{\mathbf{x\in[3,6].}}$$

Easily to see that the center of the given figure is the point $$\,\left(\dfrac92, \dfrac{11}2\right).\,$$ This allows to present it in the form of $$F(x,y)=9\left(x-\dfrac92\right)^2+2\left(x-\dfrac92\right)\left(y-\dfrac{11}2\right)+\left(y-\dfrac{11}2\right)^2-18=0.$$ or, going to main axes, $$\dfrac{2\left(\left(x-\dfrac92\right) -\big(\sqrt{17}+4\big)\left(y-\dfrac{11}2\right)\right)^2}{9(153+37\sqrt{17})} +\dfrac{2\left(\big(\sqrt{17}+4\big)\left(x-\dfrac92\right) +\left(y-\dfrac{11}2\right)\right)^2}{9(17+3\sqrt{17})}=1,$$ $$\dfrac{\left(\left(x-\dfrac92\right) -\big(\sqrt{17}+4\big)\left(y-\dfrac{11}2\right)\right)^2}{a^2} +\dfrac{\left(\big(\sqrt{17}+4\big)\left(x-\dfrac92\right) +\left(y-\dfrac{11}2\right)\right)^2}{b^2}=1.$$ where $$a=\sqrt{\frac92(153+37\sqrt{17})}\approx 37.080953,$$ $$b=\sqrt{\frac92(17+3\sqrt{17})}\approx 11.496170.$$ I.e., the given figure is a sloped ellipse.

First, we have $$0 = 9x^2+2xy+y^2-92x-20y+244 = 8(x-3)(x-6) + (x + y - 10)^2$$ which results in $$8(x - 3)(x - 6) \le 0$$. Thus, $$x \in [3, 6]$$.

Second, we have $$0 = 9x^2+2xy+y^2-92x-20y+244 = 9 (x - 5)^2 + (1 - y)^2 + (1 - y)(18 - 2x)$$ which results in $$1 - y \le 0$$ (using $$18 - 2x > 0$$). Thus, $$y \ge 1$$.

Third, we have $$0 = 9x^2+2xy+y^2-92x-20y+244 = 9(x - 4)^2 + (y - 10)^2 + 2x (y - 10)$$ which results in $$y - 10 \le 0$$ (using $$2x > 0$$). Thus, $$y \le 10$$.

Thus, $$x\in [3, 6]$$ and $$y \in [1, 10]$$.

We are done.

• The OP's question was "Where am I being wrong ?" So , your answer actually doesn't provide an answer .
– User
May 10, 2023 at 16:28
• @User I think that even the question asks a specific point, we should not reject some different solutions. Although the author asked the question, we can learn some techniques from the answers which "actually doesn't provide an answer". May 10, 2023 at 23:47

From what you found, namely $$8(x-3)(x-6) + (x+y-10)^2 = 0$$, we have $$y=10-x\pm\sqrt{8(3-x)(x-6)}.$$ Then, $$y'=-1\pm\frac{36-8x}{\sqrt{72x-8x^2-144}}=0$$ gives $$x=4$$ and $$x=5$$ as interior extrema and which in turn give the values $$y=2,10$$ and $$y=9,1$$ respectively. Hence, $$y\in[1,10].$$