Find the distance between the focii of the ellipse $(3x – 9)^2 + 9y^2 = (x\sqrt{2}+y+1)^2 $.
My approach is as follow. I have expanded the bracket
$ 9x^2-54x+81+9y^2 =2x^2+y^2+1+2\sqrt{2}xy+2y+ 2\sqrt{2}x$
$ 7x^2-54x+80+8y^2-2\sqrt{2}xy-2y-2\sqrt{2}x =0$
$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
$B^2-4AC<0$ for ellipse condition
How can I eliminate this equation , then problem becomes easy
I suspect that you’re meant to recognize that the given equation describes the ellipse in terms of a focus and directrix. Dividing both sides of the equation by $9$, it becomes $$(x-3)^2+y^2 = \frac19(x\sqrt2+y+1)^2.\tag{*}$$ The left-hand side is the square of the distance of $(x,y)$ from the point $(3,0)$, while the right-hand side is a constant times the square of the distance of $(x,y)$ from the line $x\sqrt2+y+1=0$.
So, one focus of this ellipse is at $(3,0)$, and the center of the ellipse is easily found using any of a number of standard methods; the distance between them is a straightforward computation. Alternatively, we can use the facts that the eccentricity $e$ is equal to the ratio $c/a$ of the distance between the foci and the major axis length, and that the distance of a directrix from the minor axis is equal to $a^2/c$. (Here $a$ and $c$ have their usual meanings as the semimajor axis length and half the distance between foci, respectively.) The distance between the focus and directrix is therefore $$\frac{a^2}c-c = c\left(\frac1{e^2}-1\right).$$ For this ellipse, we can extract the eccentricity $e=1/\sqrt3$ from equation (*), making the distance $2c$ between its foci equal to the distance between the focus $(3,0)$ and directrix $x\sqrt2+y+1=0$. I assume that you know how to compute that.