Find the equation of the tangent line to the curve $$2x^2-4xy+y^2-2x+6y-3=0$$ that passes through $(3,4)$.
I think I found the slope of the tangent line at some point $(x_0,y_0)$: $$y_0'=\frac{2x_0-2y_0-1}{2x_0+2y_0-6}$$ But I'm not sure what to do next.
On
Given the quadratic
$$ Q\to 2x^2-4xy+y^2-2x+6y-3=0 $$
A generic line passing by $(3,4)$ can be written as
$$ L\to \cases{x = 3+\lambda v_x\\ y = 4 +\lambda v_y} $$
now the intersections between $Q$ and $L$ can be obtained substituting $x,y$ from $L$ into $Q$ and solving for $\lambda$ or
$$ \lambda^2+2(v_y-v_x)\lambda +2v_x^2+v_y^2-4v_x v_y=0 $$
and
$$ \lambda = \frac{2(v_x-v_y)\pm 2\sqrt{v_x(7v_x-2v_y)}}{2} $$
but at tangency, the solution should be unique so
$$ v_x(7v_x-2v_y)=0\Rightarrow \cases{v_x = 0\\ v_y = \frac 72 v_x} $$
and the tangent lines are
$$ L\to \cases{x = 3\\ y = 4 +\frac 72(x-3)} $$
Attached a plot
Given conic is $$S(x,y)=2x^2-4xy+y^2-4x+6y-3=0$$ Two tangents can be drawm to a coinc from and outside point %(3,4)%, the Eq, of pairs of tangents is given as $$T^2=SS'^2,$$ Where $$T=2*3x+4y-2(4x+3y)-(x+3)+3(y+4)-3=0 \implies y-x-3=0$$, $S'=S(3,4)=4$, then the equation of pair of tangents is $$(y-x-3)^2=4(2x^2-4xy+y^2-4x+6y-3) \implies 3x^2+3y^2-10xy+2x+18y-21=0 \implies (y-3x+7)(3y-x-3)=0$$ So the Eqs. of required tangents are: $y-3x-7=0,~~~ 3y-x-3=0$