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.
Best Answer
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: $y-3x-7=0,~~~ 3y-x-3=0$