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The focus of the conic $x^{2}-6 x+4 y+1=0$ is
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Verified Answer
The correct answers are:
(3,1)
We have,
$$
x^{2}-6 x+4 y+1=0
$$
$\Rightarrow \quad(x-3)^{2}-9+4 y+1=0$
$\Rightarrow \quad(x-3)^{2}+4 y-8=0$
$\Rightarrow \quad(x-3)^{2}=-4(y-2)$
It represents parabola whose vertex is (3,2)
$\therefore$ Focus $=(3,-1+2)=(3,1)$
$$
x^{2}-6 x+4 y+1=0
$$
$\Rightarrow \quad(x-3)^{2}-9+4 y+1=0$
$\Rightarrow \quad(x-3)^{2}+4 y-8=0$
$\Rightarrow \quad(x-3)^{2}=-4(y-2)$
It represents parabola whose vertex is (3,2)
$\therefore$ Focus $=(3,-1+2)=(3,1)$
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