Question 1147397
A quadratic equation of the form {{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}} , like {{{8x^2-12xy+17y^2+16x-12y+3=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.
The value {{{B^2-4AC}}} , called the discriminant, suggests parabola, ellipse, or hyperbola if it is zero, negative, or positive respectively.
In the case of {{{8x^2-12xy+17y^2+16x-12y+3=0}}} , with {{{system(A=8,B=-12,C=17)}}} ,
{{{B^2-4AC=(-12)^2-4*8*17=144-544=-400<0}}} suggest that the equation represents an ellipse.
If there was no term in {{{xy}}} , the axes of symmetry of the ellipse would be parallel to the x- an y-axes.
If there is term in {{{xy}}} , one of the axes of symmetry will be at a positive angle {{{alpha}}} to the positive x-axis such that {{{0^o<alpha<90^o}}} .
Rotating the coordinate axes counterclockwise any angle {{{alpha}}} such that {{{0^o<alpha<90^o}}} ,
a point {{{P(x,y)}}} would be called the point {{{P(u,v)}}} with coordinates referencing the new axes,
with {{{u=x*cos(alpha)+y*sin(alpha)}}} and {{{v=-x*sin(alpha)+y*cos(alpha)}}} .
For the reverse conversion, {{{highlight(x=u*cos(alpha)-v*sin(alpha))}}} and {{{highlight(y=u*sin(alpha)+v*cos(alpha))}}}.
 
The points represented by equation {{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}} would be represented in the new set of axes
by a new equation in {{{u}}} and {{{v}}} , with new coefficients. {{{Ju^2+Kuv+Lv^2+Mu+Nv+F=0}}} .
Substituting {{{highlight(u*cos(alpha)-v*sin(alpha))}}} for{{{x}}} and {{{highlight(u*sin(alpha)+v*cos(alpha))}}} for {{{y}}} into the original equation,
we would get the equation {{{Ju^2+Kuv+Lv^2+Mu+Nv+F=0}}} .
Expressions to calculate the new coefficients can be found, including {{{K=B*(cos^2(alpha)-sin^2(alpha))+2(C-A)sin(alpha)cos(alpha))}} .
To eliminate the term in {{{uv}}} we must find a value for {{{alpha}}} that makes {{{K=0}}}
{{{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)}}} --> {{{B/(A-C)}}}{{{"="}}}{{{2sin(alpha)cos(alpha)/(cos^2(alpha)-sin^2(alpha))}}}
That can be converted into {{{highlight(B/(A-C)=sin(2alpha)/cos(2alpha)= tan(2alpha))}}} by using the trigonometric identities {{{cos(2alpha)= cos^2(alpha)-sin^2(alpha)}}} and {{{sin(2alpha)= sin(alpha)cos(alpha)}}}
From there the values for {{{2alpha}}} {{{alpha}}} and its trigonometric functions can be found.
With those values the other coefficients can be calculated as
{{{J=A*cos^2(alpha) +B*sin(alpha)cos(alpha)+C*sin^2(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)}}} .
After that, the canonical form for the equation in {{{u}}} and {{{v}}} can be found.
 
FINDING {{{alpha}}} AND ITS TRIGONOMETRIC FUNCTIONS:
As a quadratic equation of the form {{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}} , {{{8x^2-12xy+17y^2+16x-12y+3=0}}} has {{{system(A=8,B=-12,C=17,D=16,E=12,F=3)}}}
{{{B/(A-C)}}}{{{"="}}}{{{sin(2alpha)/cos(2alpha)}}}{{{"="}}}{{{tan(2alpha)}}}{{{"="}}}{{{(-12)/(8-17)}}}{{{"="}}}{{{(-12)/(-9)}}}{{{"="}}}{{{12/9}}}{{{"="}}}{{{4/3)}}}
A calculator could provide a good approximation of the value of {{{2alpha=53.13^o}}}(rounded) and that value could be used to get approximated values for {{{sin(2alpha)}}} , {{{cos(2alpha)}}} , {{{alpha}}} , {{{sin(alpha)}}} , and {{{cos(alpha)}}} .
Alternatively, exact values can be found using trigonometric identities.
Because {{{0^o<alpha<90^o}}} we know that {{{0^o<2alpha<180^o}}}}, but because {{{tan(2alpha)=4/3>0}}} we conclude {{{0^o<2alpha<90^o}}} as well, and al trigonometric functions will be positive.
From {{{sin(2alpha)/cos(2alpha)=4/3}}} <--> {{{sin(2alpha)=(4/3)cos(2alpha)}}} and trigonometric identity {{{sin^2(2alpha)+cos^2(2alpha)=1}}} we get
{{{((4/3)cos(2alpha))^2+cos^2(2alpha)=1}}} --> {{{(16/9)cos^2(2alpha)+cos^2(2alpha)=1}}} --> {{{(16/9+1)cos^2(2alpha)=1}}} --> {{{(25/9)cos^2(2alpha)=1}}} --> {{{cos^2(2alpha)=9/25}}} --> {{{cos(2alpha)=sqrt(9/25)=3/5=0.6}}} and {{{sin(2alpha)=(4/3)cos(2alpha)=(4/3)(3/5)=4/5=0.8}}} .
From {{{cos(2alpha)=3/5}}} and the trig identities for half angles we can find {{{sin^2(alpha)}}} , {{{sin(alpha)}}} , {{{cos^2(alpha)}}} , {{{cos(alpha)}}} , as
{{{sin^2(alpha)=(1-cos(2alpha))/2=(1-3/5)/2=(2/5)/2=highlight(1/5)}}} --> {{{sin(alpha)=sqrt(1/5)=sqrt(5/25)=highlight(sqrt(5)/5)}}}
{{{cos^2(alpha)=(1+cos(2alpha))/2=(1+3/5)/2=(8/5)/2=highlight(4/5)}}} --> {{{cos(alpha)=sqrt(4/5)=highlight(2sqrt(5)/5)}}}
We could also get the product {{{highlight(sin(alpha)cos(alpha)=2/5)}}} from {{{4/5=sin(2alpha)=2sin(alpha)cos(alpha)}}} --> {{{(1/2)(2sin(alpha)cos(alpha))=(1/2)(4/5)=2/5}}} .
 
CALCULATING {{{J}}}  , {{{L}}} , {{{M}}} AND {{{N}}} :
Now we can substitute the highlighted values above into the equations to calculate {{{J}}}  , {{{L}}} , {{{M}}} , and {{{N}}} :
{{{J=A*cos^2(alpha) +B*sin(alpha)cos(alpha)+C*sin^2(alpha)}}}={{{8*(4/5)+(-12)(2/5)+17*(1/5)}}}={{{32/5+17/5-24/5=25/5=5}}} ,
{{{M= D*cos(alpha)+E*sin(alpha)=16*(2sqrt(5)/5)-12(sqrt(5)/5)=(32sqrt(5)-12sqrt(5))/5=20sqrt(5)/5=highlight(4sqrt(5))=8.9443}}}{{{(rounded)}}}  , and
{{{N=-D*sin(alpha)+E*cos(alpha)=-16(sqrt(5)/5)+(-12)(2sqrt(5)/5)=
(-16sqrt(5)-24sqrt(5))/5=-40sqrt(5)/5=highlight(-8sqrt(5))=-17.8886}}}{{{(rounded)}}} .
 
FINDING THE CANONICAL FORM AND THE NATURE OF THIS CONIC:
For now, substituting the values found for {{{J}}} and {{{L}}} , we can write the equation of the conic in terms of {{{u}}} and {{{v}}} as:
{{{5u^2+20v^2+Mu+Nv+3=0}}} ,
and we can and need to transform it into something of the form {{{(u-h)^2/p^2+(v-k)^2/q^2=1}}} which would be equivalent to
{{{((u-h)/p)^2+((v-k) /q)^2=1}}} .
That is the equation of a circle of radius {{{1}}} , stretched in the u and v directions by factors {{{p}}} and {{{q}}} , centered in point (h,k) .
In other words, that represents an ellipse centered in (h,k) . 
To get that transformation we need to form the squares {{{(u-h)^2=u-2uh+h^2}}} and {{{+(v-k)^2=v-2vk+k^2}}} from {{{5u^2+Mu}}} and {{{20v^2 +Nv}}} respectively .
{{{5u^2+20v^2+Mu+Nv+3=0}}} --> {{{5u^2+Mu+20v^2+Nv=-3}}} --> {{{5(u^2+2(M/10)u+(M/10)^2)+20(v^2+2(N/40)v+(N/40)^2)=-3+5(M/10)^2+20(N/40)^2}}} --> 
{{{highlight(5(u+ (M/10))^2+20(v+(N/40))^2=-3+5(M/10)^2+20(N/40)^2)}}} 
{{{M/10=4sqrt(5)/10=highlight(0.4sqrt(5))=0.89443}}}{{{(rounded)}}} ,
and {{{N/40=-8sqrt(5)/40=highlight(-0.2sqrt(5))=-0.447215}}}{{{(rounded)}}}  .
{{{(M/10)^2=(0.4sqrt(5))^2=0.16*5=0.8}}} 
{{{(N/40)^2=(-0.2sqrt(5))^2=0.04*5=0.2}}}
Then, the right hand side term is {{{-3+5(M/10)^2+20(N/40)^2}}}={{{-3+5*(0.4sqrt(5))^2+20*(0.5sqrt(5))^2}}}={{{-3+5*0.16*5+20*0.04*5=-3+4+4=5}}} , and the equation is
{{{highlight(5(u+0.4sqrt(5))^2+20(v+(-0.2sqrt(2)))^2=5)}}}
From there, we continue:
{{{5(u+0.4sqrt(5))^2+20(v-0.2sqrt(5))^2=5}}} --> {{{(u+0.4sqrt(5))^2+4(v-0.2sqrt(2))^2=1}}} --> 
{{{(u-(-0.4sqrt(5)))^2/1^2}}}{{{"+"}}}{{{(v-0.2sqrt(5))^2/0.5^2}}}{{{"="}}}{{{1}}}
The center of the ellipse is at {{{"("}}}{{{-0.4sqrt(5)}}}{{{","}}}{{{0.2sqrt(5)}}}{{{")"}}} .

The numbers squared in the denominators are the semi-major and semi-minor axes of the ellipse.
The greater one, {{{1}}} , is the semi-major axis, and the other I the semi-minor axis
The extreme values for {{{u}}} happen when {{{v-0.2sqrt(5)=0}}} , and then
{{{(u+0.4sqrt(5))^2=1}}} --> {{{u+0.4sqrt(5)=" "+- 1}}} --> {{{u=-0.4sqrt(5) +- 1}}} .
The y-coordinate of the center and the extremes show the vertices are at a distance of {{{1}}} from the center .
Using the equations {{{highlight(x=u*cos(alpha)-v*sin(alpha))}}} and {{{highlight(y=u*sin(alpha)+v*cos(alpha))}}} , we can find the coordinates of  the center and vertices of the ellipse.
For the center, {{{x=(-0.4sqrt(5))*(2sqrt(5)/5)-0.2sqrt(5)*(sqrt(5)/5)}}}=
{{{-0.8*5/5-0.04sqrt(25)=-0.8-0.04*5=-0.8-0.2=-1}}}
{{{y=-0.4sqrt(5)*(sqrt(5)/5)+0.2sqrt(5)*(2sqrt(5)/5)}}}=
{{{-0.4*5/5+0.2*2*5/5=-0.4+0.4=0}}}



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