Deal with non standard form of conic I want to know how can I calculate latus rectum, tangent at vertex, vertex and axes of a parabola whose equation is not standard. For example, the parabola:
$$
4x^2 - 4xy + y^2 - 10 y - 19 = 0
$$
 A: The procedure of dealing with arbitrary $ax^2+bxy+cy^2+dx+ey+f=0$ is first translation then rotation.
First move the origin of coordinate system from $(0,0)$ to $(h,k)$, we have $(x,y)=(x'+h,y'+k)$ where $x'$ and $y'$ denote the new coordinate.
We'll get the new equation $a(x')^2+bx'y'+c(y')^2+(2ah+bk+d)x'+(bh+2ck+e)y'+(ah^2+bhk+ck^2+dh+ek+f)=0$
To pick proper $(h,k)$ to make the linear term disappear solve $2ah+bk+d=0$ and $bh+2ck+e=0$. We have(if $b^2-4ac \neq 0$)
$$h=\frac{2cd-be}{b^2-4ac},k=\frac{2ae-bd}{b^2-4ac}$$. If $b^2=4ac$ you can only make one linear term disappear, the graph is usually parobala.
Now you have $a(x')^2+bx'y'+c(y')^2+f'=0$, we rotate the coordinate system(counter-clockwise), by $\theta$. We have $(X,Y)=(x'\cos\theta+y'\sin\theta,-x'\sin\theta+y'\cos\theta)$, the new equation becomes $(a\cos^2\theta+b\sin\theta\cos\theta+c\sin^2\theta)X^2+((c-a)\sin2\theta+b\cos2\theta)XY+(c\cos^2\theta-b\sin\theta\cos\theta+a\sin^2\theta)Y^2+f'=0$
To pick the correct angle to rotate to make $XY$ term disappear, choose
$$\cot(2\theta)=\frac{a-c}{b}$$
After you've done all the work, it'll become standard form, and you can find out all the properties you want.
