Inclined parabola Please help me to solve the inclined parabola.
Is it that we have to convert this entire equation in the form of axis  and tangent at vertex. That would be tiresome. Any other method?
For this  parabola,$x^2 +y^2 +2xy -6x -2y + 3=0$,How can we find it's focus?
 A: Since you don't like the easy way (see my comment), let's do it the hard way: rotate $x^2+2xy+y^2-6x-2y+3=0$ around the origin by $\frac{\pi}{4}$: 
$$x=\frac{x'-y'}{\sqrt{2}}$$
$$y=\frac{x'+y'}{\sqrt{2}}$$
substituting this and expanding the equation reduces to:
$$y'+\frac{x'^2}{\sqrt{2}}-2x'+\frac{3}{2\sqrt{2}}=0$$ or 
$$y'-\frac1{2\sqrt{2}}=-\frac{(x'-\sqrt{2})^2}{\sqrt{2}}$$ or
$$y''=-\frac{x''^2}{\sqrt{2}}=\frac1{4f}x''^2$$ which makes $f=-\frac1{2\sqrt{2}}$ and the focus for this parabola is $(x'',y'')=(0,f)$. Now we go back through the transformations and get $(x',y')=(\sqrt{2},0)$ an the inverse rotation takes this to $(x,y)=(1,1)$.
A: Let There be a parabola $(ax+by)^2+2gx+2fy+c=0 $
This can be written as $(ax+by)^2=-2gx-2fy-c$
Now we will add arbitrary constant in square term,
$$(ax+by+\lambda)^2=xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)-(1)$$
Now simply choose $\lambda$ such that the lines,
$ax+by+\lambda$ and $xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)$
are perpendicular.
now $xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)=bx-ay+\mu$
we can write (1) as,
$$(\frac{ax+by+\lambda}{\sqrt{a^2+b^2}})^2=4\rho(\frac{bx-ay+\mu}{\sqrt{a^2+b^2}})$$
which is of the form $Y^2=4\rho X$
Now we can do our stuff on this reduced form,
1)LATUSRECTUM:- $4\rho$=$\frac{1}{\sqrt{a^2+b^2}}$
2)AXIS:- Y=0 or ax+by+$\lambda$=0
3)EQUATION OF TANGENT AT VERTEX:- X=0 or bx-ay+$\mu$=0
4)VERTEX:- Intersection of Y=0 and X=0
5)EQUATION OF DIRECTRIX:- X+$\rho$=0
6)EQUATION OF LATUS RECTUM:- X-$\rho$=0
7)FOCUS:- By solving X-$\rho$=0 and X=0
well now we can apply all the basic parabola properties in short {^w^} .
