2x2-3xy+2y2-2=0.
Treat y as a constant and solve for x using the quadratic formula:
x=(3y±√(9y2-8(2y2-2))/4=(3y±√(16-7y2))/4.
This enables irrational factorisation:
(x-¾y+√(16-7y2)/4)(x-¾y-√(16-7y2)/4)=0.
Because there's only one equation but two variables this is the best that can be done---relating x and y.
Note that when y=1, x=¾±¾, so x=0 or 3/2; when y=-1, x=-¾±¾, so x=0 or -3/2; also when y=0, x=±1.
If the equation is plotted as a graph we get a tilted (45°) ellipse (y=x and y=-x form the orthogonal axes), which passes through (1,0), (-1,0), (3/2,1),(-3/2,-1). The equation of the rectified ellipse is x2+7y2=4.
DERIVATION OF THE EQUATION OF THE ELLIPSE
Consider a point P(x,y) on the graph of the given equation. The origin is O(0,0) and OP can be represented by r, making P(rcosφ,rsinφ) where φ is the angle between OP and the x-axis.
If we now introduce new axes X and Y such that the X-axis is inclined at angle θ to the x-axis, then the Y-axis will also be inclined to the y-axis by the same angle θ (measured anticlockwise).
The angle between OP and the X-axis is φ-θ, making P(rcos(φ-θ),rsin(φ-θ)). So X=rcos(φ-θ) and Y=rsin(φ-θ).
Using trig identities: X=rcos(φ-θ)=rcosφcosθ+rsinφsinθ, Y=rsinφcosθ-rcosφsinθ. Now substitute x for rcosφ and y for rsinφ:
X=xcosθ+ysinθ, Y=ycosθ-xsinθ. We need x and y in terms of X and Y, solving this system of equations we get:
Xsinθ+Ycosθ=y(sin2θ+cos2θ)=y; Xcosθ-Ysinθ=x(cos2θ+sin2θ)=x. Now we can substitute for x and y in the given equation:
2(Xcosθ-Ysinθ)2-3(Xcosθ-Ysinθ)(Xsinθ+Ycosθ)+2(Xsinθ+Ycosθ)2-2=0.
This looks complicated and frightening! But we're going to use it to get rid of that troublesome xy term. We need to find the value of θ which will make the XY term become zero. So we only need to look at the contributions to the XY term. The first contribution comes from the first squared term 2(Xcosθ-Ysinθ)2. This yields -4XYsinθcosθ. The next term yields -3XY(cos2θ-sin2θ); the last term yields 4XYsinθcosθ.
When we add these we get only -3XY(cos2θ-sin2θ)=-3XYcos(2θ), which is zero when cos(2θ)=0, 2θ=π/2, θ=π/4. sinθ=cosθ=1/√2 when θ=π/4. So the original equation becomes:
2(X/√2-Y/√2)2-3(X/√2-Y/√2)(X/√2+Y/√2)+2(X/√2+Y/√2)2-2=0,
2(X-Y)2/2-3(X-Y)(X+Y)/2+2(X+Y)2/2-2=0,
(X-Y)2-3(X2-Y2)/2+(X+Y)2-2=0,
X2-2XY+Y2-3X2/2+3Y2/2+X2+2XY+Y2-2=0,
X2/2+7Y2/2=2, X2+7Y2=4 or X2/4+7Y2/4=1 in standard form.
In the X-Y frame of reference, the relation between X and Y is simpler. Geometrically the ellipse has a semi-major axis length 2 and a semi-minor axis length 2/√7.