Focus of the Parabola Find the Focus of $$(2x+y-1)^2=5(x-2y-3)$$.
Clearly its a Parabola whose axis is $2x+y-1=0$ and since $x-2y-3=0$ is perpendicular to $2x+y-1=0$ Tangent at the vertex is $x-2y-3=0$.Also the Vertex is $(3,-1)$, but now how to find its focus?

More generally if we have a parabola in the form (which all parabolas can be put into) $$(l_{\rm axis})^2=e\cdot l_{\text{tangent at vertex}}$$ or $$(ax+by+c)^2=e(-bx+ay+d),$$ and use the orthonormal change of basis $x'=\frac{ax+by}{\sqrt{a^2+b^2}},y'=\frac{-bx+ay}{\sqrt{a^2+b^2}}$, we get $$(\sqrt{a^2+b^2}x'+c)^2=(\sqrt{a^2+b^2}y'+d)$$ or $$(a^2+b^2)((x'+\frac{c}{\sqrt{a^2+b^2}})^2-\frac{e}{\sqrt{a^2+b^2}}(y'+\frac{d}{\sqrt{a^2+b^2}}))=0.$$ So using that$$x''^2=4\frac{e}{4\sqrt{a^2+b^2}}y''$$ has focus $$(x'',y'')=(0,\frac{e}{4\sqrt{a^2+b^2}})$$ we nest back to $$(x',y')=(-\frac{c}{\sqrt{a^2+b^2}},\frac{e-4d}{4\sqrt{a^2+b^2}})$$ and get $$(x,y)=(-\frac{a(4d-e)+4bc}{4b^2+4a^2},-\frac{(b(e-4d)+4ac}{4b^2+4a^2}).$$
In your special case the focus is $(x,y)=(\frac54,-\frac32).$
