Question 1158804
A quadratic equation of the form {{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}} , like {{{9x^2-24xy+16y^2-20x-15y-50=0}}} , could represent a circle, ellipse, hyperbola, parabola.
In special cases it could represent a point, a line, a pair of lines, or no point that could exist, depending on the values of the coefficients.
 
ROTATING AXES:
Axis rotation equations:
From coordinates (x, y), to get the new coordinates (u, v) that would result for a point (x,y) after rotating the coordinate axes counterclockwise an angle {{{ alpha }}} we can apply
{{{u=x*cos(alpha)+y*sin(alpha)}}}
{{{v=-x*sin(alpha)+y*cos(alpha)}}}
From (u, v), rotating an angle {{{-alpha}}} (the reverse change) we get (x,y) using
{{{x=u*cos(-alpha)-v*sin(-alpha)}}} --> {{{highlight(x=u*cos(alpha)-v*sin(alpha))}}}
{{{y=-u *sin(-alpha)+v*cos(-alpha))}}} --> {{{highlight(y=u*sin(alpha)+v*cos(alpha))}}}
Substituting {{{u*cos(alpha)-v*sin(alpha)}}} for {{{x}}} and {{{ u*sin(alpha)+v*cos(alpha)}}}  for {{{y}}} in {{{9x^2-24xy+16y^2-20x-15y-50=0}}} ,
with a lot of busy algebra work, we could get a new equation in {{{u}}} and {{{v}}} with new expressions for the new coefficients.
That is grueling, mistake-inducing work (and torture to type and edit).
We would have expressions involving {{{alpha}}} for all the new coefficients of terms with {{{u}}} and/or {{{v}}} , including the coefficient of the term in {{{uv}}} .
That coefficient must be made equal to zero.
From the equation making the coefficient of {{{uv}}} equal to zero, a value for {{{alpha}}} needs to be found to be used to calculate all other coefficients.
Maybe that is the initial work that was expected for this problem.
 
IDENTIFYING VERTEX (OR VERTICES):
The new equation on {{{u}}} and {{{v}}} should be identified as a parabola (one vertex), a hyperbola (two vertices), an ellipse (two vertices and two co-vertices), or something else.
Then, the (u,v) coordinates for any vertices should be found, and converted into (x,y) coordinates.
 
POSSIBLE SHORTCUTS:
Rotating the coordinate axes counterclockwise an angle {{{alpha}}} , from an equation 
{{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}} we get a new equation in {{{u}}} and {{{v}}} with new coefficients:
{{{Jx^2+Kxy+Ly^2+Mx+Ny+F=0}}} .
The new coefficients can be found to be:
{{{J=A*cos^2(alpha) +B*sin(alpha)cos(alpha)+C*sin^2(alpha) }}} ,
{{{K= B*(cos^2(alpha)-sin^2(alpha))+2(C-A) sin(alpha)cos(alpha)})}} ,
{{{L= A*sin^2(alpha) -B*sin(alpha)cos(alpha)+C*cos^2(alpha)}}} ,
{{{M= D*cos(alpha)+E*sin(alpha)}}} , and {{{N=-D* sin(alpha)+E*cos(alpha)}}} . 
To eliminate the term in {{{uv}}} we must find a value for {{{alpha}}} that makes
{{{B*(cos^2(alpha)-sin^2(alpha))+2(C-A) sin(alpha)cos(alpha)=0})}} {{{B*(cos^2(alpha)-sin^2(alpha))=2(A-C)sin(alpha)cos(alpha)})}} .
That equation above can be re-written as a simpler one based on angle {{{2alpha}}} using the trigonometric identities
{{{sin(2alpha)=2 sin(alpha)cos(alpha)})}}and {{{cos(2alpha)= cos^2(alpha)-sin^2(alpha)}}}
Then, it becomes {{{B*cos(2alpha)=(A-C)*sin(2alpha)}}} --> {{{ B/(A-C)=sin(2alpha)/ cos(2alpha)}}} --> {{{system(B/(A-C)=X, sin(2alpha)=cos(2alpha)(B/(A-C)))}}} . B/(A-C)=tan(2alpha), sin(2alpha)=cos(2alpha)(B/(A-C)
Avoiding mistakes, with careful algebra work, I might have been able to find the expressions outlined above.
I just copied them from an old Analytical Geometry textbook.
The value {{{B^2-4AC}}} , called the discriminant, suggests parabola, ellipse, or hyperbola if it is zero, negative, or positive respectively.
I got  that from the book too.
In the case of {{{9x^2-24xy+16y^2-20x-15y-50=0}}} ,
{{{B^2-4AC=24^2-4*9*16=3^2*8^2-9*8^2=576-576=0}}} says the equation represents a parabola.
 
A SHORTCUT THAT SHOULD BE ALLOWED:
Using {{{system(x=u*cos(alpha)-v*sin(alpha), y=u*sin(alpha)+v*cos(alpha))}}} , we can translate into (u, v) coordinates the terms of degree 2 (the first 3 terms):
{{{9x^2=9(u*cos(alpha)-v*sin(alpha))^2}}}{{{"="}}}{{{ red(9u^2*cos^2(alpha))-18uv*sin(alpha)cos(alpha)+blue(9v^2*sin^2(alpha))}}}
 
{{{-24xy=-24(u*cos(alpha)-v*sin(alpha))(u*sin(alpha)+v*cos(alpha))
}}}{{{"="}}}{{{-24(u^2*cos(alpha)sin(alpha) +uv*cos^2(alpha) -uv*sin^2(alpha) -v^2*sin(alpha)cos(alpha))}}}{{{"="}}}{{{red(-24u^2*cos(alpha)sin(alpha))-24uv*cos^2(alpha)+24uv*sin^2(alpha)+blue(24v^2*sin(alpha)cos(alpha))}}}{{{"="}}}{{{red(-24u^2*cos(alpha)sin(alpha))-24uv(cos^2(alpha)-sin^2(alpha))+blue(24v^2*sin(alpha)cos(alpha))}}}
 
{{{16y^2=16(u*sin(alpha)+v*cos(alpha))^2}}}{{{"="}}}{{{red(16u^2* sin^2(alpha))+32uv*sin(alpha)cos(alpha)+blue(16v^2*cos^2(alpha)))}}} 
 
When “collecting like terms”, the term in {{{uv}}} will be
{{{-18uv*sin(alpha)cos(alpha) +32uv*sin(alpha)cos(alpha)}}} . 
Making the coefficient of {{{uv}}} equal to zero, we get the equation
{{{-18sin(alpha)cos(alpha) +32sin(alpha)cos(alpha)=0}}} 
We can simplify that equation using the trigonometric identities
{{{cos(2alpha)=cos^2(alpha)-sin^2(alpha)}}} and {{{sin(2alpha)=2sin(alpha)cos(alpha)}}} .
We get {{{7sin(2alpha)=24cos(2alpha)}}} --> {{{system(sin(2alpha)=(24/7)cos(2alpha), and, tan(2alpha)=24/7)}}} From that we would not get an exact value for {{{2alpha}}} or {{{alpha}}} , but we know that alpha is in.
A calculator would give you an approximation with enough digits to figure out that {{{cos(alpha)=0.8}}} and {{{sin(alpha)=0.6}}} .
We can get exact values of the trigonometric functions and their squares, from {{{sin(2alpha)=(24/7)cos(2alpha)}}}
{{{sin^2(2alpha)+cos^2(2alpha)=1}}} --> {{{((24/7)cos(2alpha))^2+cos^2(2alpha)=1}}} --> {{{(576/49)cos^2(2alpha)+cos^2(2alpha)=1}}} --> {{{(576/49+1)cos^2(2alpha)=1}}} --> {{{(625/49)cos^2(2alpha)=1}}} --> {{{cos^2(2alpha)=49/625}}} --> {{{highlight(cos(2alpha)=7/25)}}} and {{{highlight(cos(2alpha)=7/25=cos^2(alpha)-sin^2(alpha))}}}
Then, {{{sin(2alpha)=(24/7)cos(2alpha)}}} --> {{{sin(2alpha)=(24/7)(7/25)=24/25}}} , so {{{highlight(sin(2alpha)=24/25=2sin(alpha)cos(alpha))}}}
The sine and cosine of {{{alpha}}} can also be calculated from {{{cos(2alpha)}}} using the trigonometric identities for half angles as
{{{sin(alpha)=sqrt((1-cos(2alpha))/2)=sqrt((1-7/25)/2)=sqrt(9/25)=3/5=0.6}}} --> {{{highlight(sin(alpha)=3/5=0.6)}}} and
{{{cos(alpha)=sqrt((1+cos(2alpha))/2)=sqrt((1+7/25)/2)=sqrt(16/25)=4/5=0.8}}} --> {{{highlight(cos(alpha)=4/5=0.8)}}}
 
Now we can go back to expressions for the terms in {{{u^2}}} and {{{v^2}}} , and substitute for the trigonometric functions involved the values we found and highlighted before.
As found before they come from the original terms {{{9x^2}}} , {{{-24xy}}} , and {{{16y^2}}} expressed as a function of {{{u}}}, {{{v}}} and {{{alpha}}} : 
{{{highlight(sin(2alpha)=24/25-sin(alpha)cos(alpha))}}} ,
{{{highlight(cos(2alpha)=7/25=cos^2(alpha)-sin^2(alpha))}}} , 
{{{highlight(sin(alpha)=3/5=0.6)}}} and {{{highlight(cos(alpha)=4/5=0.8)}}}
{{{9x^2=9(u*cos(alpha)-v*sin(alpha))^2}}}{{{"="}}}{{{ red(9u^2*cos^2(alpha))-18uv*sin(alpha)cos(alpha)+blue(9v^2*sin^2(alpha))}}}
{{{-24xy}}}{{{"="}}}{{{red(-24u^2*cos(alpha)sin(alpha))-24uv(cos^2(alpha)-sin^2(alpha))+blue(24v^2*sin(alpha)cos(alpha))}}} ,
and
{{{16y^2=16(u*sin(alpha)+v*cos(alpha))^2}}}{{{"="}}}{{{red(16u^2* sin^2(alpha))+32uv*sin(alpha)cos(alpha)+blue(16v^2*cos^2(alpha)))}}}
Collecting like terms, we find the term in {{{u^2}}} to be
{{{red(9u^2*cos^2(alpha))-red(24u^2*cos(alpha)sin(alpha))+red(16u^2* sin^2(alpha))}}}{{{"="}}}{{{red(9(4/5)^2(alpha)-24(4/5)(3/5)+16*(3/5)^2)u^2}}}{{{"="}}}{{{red(9*16/25-12(12/25)+16*(9/25))u^2}}}{{{"="}}}{{{red((144-288-1440/25))u^2}}}{{{"="}}}{{{highlight(red(0)*u^2)}}} ,
and the tern in {{{v^2}}} to be
{{{blue(9v^2*sin^2(alpha))+blue(24v^2*sin(alpha)cos(alpha))+blue(16v^2*cos^2(alpha))}}}{{{"="}}}{{{blue(9(3/5)^2+24(3/5)(4/5)+16(4/5)^2)v^2}}}{{{"="}}}{{{blue(9*9/25+24*12/25+16(16/25))v^2}}}{{{"="}}}{{{blue(81+288+256/25))v^2}}}{{{"="}}}{{{blue(625/25))v^2}}}{{{"="}}}{{{highlight(blue(25)v^2)}}}
 
The linear (degree 1) terms in {{{u}}} and {{{v}}} come from
{{{-20x=-20(u*cos(alpha)-v*sin(alpha))}}}{{{"="}}}{{{-20((4/5)u-(3/5)v)=-16u+12v}}} , and
{{{-15y=-15(u*sin(alpha)+v*cos(alpha))}}}{{{"="}}}{{{-15((3/5)u+(4/5v))=-9u-12v}}}
{{{-20x-15v=-16u+12v-9u-12v=-25u+0v}}}
The equation on {{{u}}} and {{{v}}} is
{{{red(0)*u^2+0*uv+blue(25)v^2-25u+0v-50=0}}} or {{{25v^2-25u-50=0}}} , which simplifies to {{{v^2-u-2=0}}} or {{{u=v^2-2}}} , representing a parabola with axis {{{v=0}}}, and vertex {{{V(-2,0)}}} all in u,v coordinates.
The x,y coordinates of the vertex are
{{{x=u*cos(alpha)-v*sin(alpha)=-2*(4/5)-0*(3/5)=-8/5=-1.6}}}
{{{y=u*sin(alpha)+v*cos(alpha)=-2*(3/5)+0*(4/5)=6/5=-1.2}}}