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  4. The focus of the conic x2 - 6x + 4y + 1 = 0 is

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

WBJEEWBJEE 2017

Solution:

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