Find the point to which origin should be shifted to reduce the equation $$3x^2 - 2xy + 4y^2 + 8x - 10y + 8 = 0$$ into one with linear term missing.
So I just know that a linear term is where the power of $x$ is 1 but besides that I have no idea how to solve this. I tried solving it by factorisation and also by trying to change it to $(a+b)^2$ or $(a-b)^2$ formats but it just kept getting confusing. Please help.
A quadratic equation with no linear term gives a curve which is centered on the origin (as is evidenced by $180^\circ$ rotational symmetry about the origin; swap $x$ with $-x$ and $y$ with $-y$, and if there are no linear terms, we get exactly the same equation back). So finding the center of the ellipse and putting our new origin there seems like it ought to work.
The center of the ellipse can be found by finding its bounding rectangle (with sides parallel to the axes), and the sides of that rectangle are found by solving the equation with respect to either $x$ or $y$ and checking when that equation has exactly one solution. In fact, let's just find the discriminant $\Delta$ immediately and solve $\Delta = 0$.
Solving for $x$ we get $$ 3x^2 + (8 - 2y)x + 4y^2 -10y +8= 0\\ \Delta_x = (8 - 2y)^2 - 4\cdot 3 \cdot(4y^2 -10y+8) = 0\\ 64 - 32y + 4y^2 - 48y^2 + 120y - 96 = 0\\ y = 1\pm\sqrt{\frac3{11}} $$ The middle point between these is just $y = 1$.
And solving for $y$, we get $$ 4y^2 + (-10-2x)y + 3x^2 + 8x + 8 = 0\\ \Delta_y = (-10-2x)^2 - 4\cdot 4\cdot(3x^2 + 8x + 8) = 0\\ x = -1\pm\sqrt{\frac4{11}} $$ with the middle point $x = -1$.
So it seems like the center of the ellipse is at $(-1, 1)$. Let's make a new $uv$-coordinate system with this as the origin. This gives $x = u-1$ and $y = v+1$, which we can just insert: $$ 3(u-1)^2 - 2(u-1)(v+1) + 4(v+1)^2 + 8(u-1) - 10(v+1) + 8 = 0\\ 3u^2 - 6u + 3 - 2uv - 2u + 2v + 2 + 4v^2 + 8v + 1 + 8u-8 -10v - 10 + 8 = 0\\ 3u^2 + 3 - 2uv+ 4v^2 - 7 = 0 $$ and we have found a new coordinate system where the equation for the ellipse has no linear term.