rotate the aces to eliminate the xy term for 2x^2-72xy+23y^2-80x-60y-125+0

I think this is supposed to be :

2x^2-72xy+23y^2-80x-60y-125=0

We need to first calculate this:

cot2Θ = [A - C]  / B   = [2 - 23]/[-72]  = [-21]/[-72]  = [7/24]

And, from trig, we have that cos2Θ  = 7/25

And using trig again, we can caclulate  sinΘ  and cosΘ, thusly

 sinΘ  = √[(1 - cos2Θ)/2 ] = √[(1 - 7/25)/2] = √[9/25] = 3/5

cosΘ =  √[(1 + cos2Θ)/2 ] = √[(1 + 7/25)/2] = √[16/25] = 4/5

Now....we need to make the following substitutions

x = cosΘ X  -  sinΘ Y      and y =  sinΘ X + cosΘ Y    ....  or, just .......

x = (4/5)X - (3/5)Y       and  y = (3/5)X + (4/5)Y

So we have

2[(4/5)X - (3/5)Y]^2-72[ (4/5)X - (3/5)Y][(3/5)X + (4/5)Y]+23[(3/5)X + (4/5)Y]^2-80[ (4/5)X - (3/5)Y]-60[ (3/5)X + (4/5)Y -125=0-80[ (4/5)X - (3/5)Y]  

And term by term, we have.......

[(32/25)X^2  - (48/25)XY + (18/25)Y^2] +

[(-864/25)X^2 - (504/25)XY + (864/25)Y^2] +

[(207/25)X^2 +(552/25)XY + (368/25)Y^2]  +

[48Y  - 64X]  +

[-36X  - 48Y ] -

125    =  0

Simplifying the above, we get

50Y^2  - 25X^2 -100X  - 125 = 0     divide through by 25

2Y^2 - x^2 - 4x - 5 = 0   complete the square on x  

2Y^2 - [x^2 + 4x + 4  - 4] = 5

2Y^2 - (x +2)^2 = 1

And we have the recognizable equation of a hyperbola.

Here's a picture of both the original  graph with the xy term and the "normal" graph without rotation...

https://www.desmos.com/calculator/rrkgvzswml

BTW...the original graph's rotation can be found thusly:

cot2Θ = [A - C]  / B   = [2 - 23]/[-72]  = [-21]/[-72]  = [7/24] 

cot^-1(7/24) =73.739795291688°

So half of this is the angle of rotation ≈ 36.87° 

$$\mathrm{\ }$$Could you repeat that in English? Thank You!

Very impressive Chris    

Oh, rotate the Axes  at least the question makes some degree of sense.

I will have to study your answer Chris ://

I would have had NO idea how do do this one!

Melody....I only vaguely remembered this from Calculus.....I had to do some "reviewing," too.  I don't know if this is taught much now in many places. The algebra gets pretty "sticky".......

Rotate the axis to eliminate the xy term for 2x^2-72xy+23y^2-80x-60y-125=0

$$\small{\text{$
\begin{array}{ccccccccccc}
\textcolor[rgb]{1,0,0}{2}\cdot x^2 &+& \textcolor[rgb]{1,0,0}{23} \cdot y^2 &-& \textcolor[rgb]{1,0,0}{72} \cdot x \cdot y &-& \textcolor[rgb]{1,0,0}{80} \cdot x &-& \textcolor[rgb]{1,0,0}{60} \cdot y &=& 125\\
\textcolor[rgb]{1,0,0}{a} \cdot x^2 &+& \textcolor[rgb]{1,0,0}{b} \cdot y^2 &+& \textcolor[rgb]{1,0,0}{2\cdot c} \cdot x \cdot y &+& \textcolor[rgb]{1,0,0}{2\cdot d} \cdot x &+& \textcolor[rgb]{1,0,0}{2\cdot e} \cdot y &=& 125\\
\end{array}
$}}$$

$$\\
\small{\text{$
\begin{array}{rcr}
a &=& 2 \\
b &=& 23 \\
c &=& -36 \\
d &=& -40 \\
e &=& -30
\end{array}
$}}$$

$$\boxed{ \small{\text{$
\tan{(\varphi)}=\dfrac{ \sqrt{ (a-b)^2+4\cdot c^2 }-(a-b) } {2\cdot c}
$}} }\\\\
\small{\text{$
\tan{(\varphi)}=\dfrac{ \sqrt{ (2-23)^2+4\cdot (-36)^2 }-(2-23) } {2\cdot (-36) }
= \dfrac{ \sqrt{ (-21)^2+4\cdot (-36)^2 }+21 } { -72 }
$}}\\\\
\small{\text{$
\tan{(\varphi)}
= \dfrac{ \sqrt{ 441+4\cdot 1296 }+21 } { -72 }
$}}\\\\
\small{\text{$
\tan{(\varphi)}
= \dfrac{ 96 } { -72 }
$}}\\\\
\small{\text{$
\tan{(\varphi)}
= - \dfrac{ 4 } { 3 }
$}}\\$$

$$\boxed{
\small{\text{
$
\sin{ (\varphi) }
=\dfrac { \tan{(\varphi)} }
{ \sqrt{1+\tan{(\varphi)}^2 } } \qquad
\cos{ (\varphi) }
=\dfrac { 1 }
{ \sqrt{1+\tan{(\varphi)}^2 } }
$}}
}\\\\
\small{\text{
$
\sin{ (\varphi) }
=\dfrac { -\frac{4}{3} }
{ \sqrt{1+ (-\frac{4}{3} )^2 } } \qquad
\cos{ (\varphi) }
=\dfrac { 1 }
{ \sqrt{1+ (-\frac{4}{3} )^2 } }
$}}\\\
\small{\text{
$
\sin{ (\varphi) }
=-\dfrac { 4 } { 5 } \qquad
\cos{ (\varphi) }=\dfrac { 3 } { 5 }
$}}$$

Rotation: $$\boxed{
\small{\text{
$
\begin{array}{rcl}
x &=& x'\cdot \cos{(\varphi)}-y'\cdot \sin{(\varphi)} \\
y &=& x'\cdot \sin{(\varphi)}+y'\cdot \cos{(\varphi)}
\end{array}
$}}}$$

$$\small{\text{
$
\begin{array}{rcl}
x &=& x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5} \\
y &=& -x'\cdot \dfrac{4}{5}+y'\cdot \dfrac{3}{5}
\end{array}
$}}$$

We substitute:

$$\\\small{\text{
$
x^2 = \left(
x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5}
\right)^2
= \frac{9}{25}x'^2 + 2 \frac{12}{25}x'y'+\frac{16}{25}y'^2
$}}\\
\small{\text{
$
y^2 = \left(
-x'\cdot \dfrac{4}{5}+y'\cdot \dfrac{3}{5}
\right)^2
=\frac{16}{25}x'^2 - 2 \frac{12}{25}x'y'+\frac{9}{25}y'^2
$}}\\
\small{\text{
$
xy =\left(
x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5}
\right)\cdot
\left(
-x'\cdot \dfrac{4}{5}+y'\cdot \dfrac{3}{5}
\right)
= -\frac{12}{25}x'^2 + \frac{12}{25}y'^2 - \frac{7} {25} \cdot x'y'
$}}\\$$

$$\small{\text{$
\begin{array}{rcl}
2\cdot (\frac{9}{25}x'^2 + 2 \frac{12}{25}x'y'+\frac{16}{25}y'^2) \\\\
+ 23 \cdot (\frac{16}{25}x'^2 - 2 \frac{12}{25}x'y'+\frac{9}{25}y'^2) \\\\
-72 \cdot (-\frac{12}{25}x'^2 +\frac{12}{25}y'^2 - \frac{7}{25} \cdot x'y') \\\\
- 80 \cdot ( x'\cdot \dfrac{3}{5}+y'\cdot \dfrac{4}{5} ) \\\\
-60 \cdot (-x'\cdot \dfrac{4}{5}-y'\cdot \dfrac{3}{5}) &=& 125
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\frac{18}{25}x'^2 +\frac{32}{25}y'^2 + \frac{368}{25}x'^2 +\frac{207}{25}y'^2+\frac{864}{25}x'^2 -\frac{864}{25}y'^2
-\frac{240}{5} x' - \frac{320}{5} y'
+\frac{240}{5} x'+\frac{180}{5} y' &=& 125\\\\
50\cdot x'^2 -25\cdot y'^2 +0 \cdot x' -28\cdot y' &=& 125\\\\
50\cdot x'^2 -25\cdot y'^2 -28\cdot y' &=& 125
\end{array}
$}}$$