Question

The focus of the conic $x^2-6x+4y+1=0$ is

• A. $(2,3)$
• B. $(3,2)$
• C. $(3,1)$
• D. $(1,4)$

Answer 1

Answer: (C)

We have,
$x^2-6x+4y+1=0$
$\Rightarrow (x-3{)}^2-9+4y+1=0$
$\Rightarrow (x-3{)}^2+4y-8=0$
$\Rightarrow (x-3{)}^2=-4(y-2)$
It represents parabola whose vertex is $(3,2)$
$\therefore$ Focus $=(3,-1+2)=(3,1)$