The axis of the conic $x^{2} + 4 y - 6 x + 17 = 0$ is

x = 5

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y = 5

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x = 3

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x = - 3

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Solution

The correct option is A $x = 3$
$x^{2} + 4 y - 6 x + 17 = 0$
$\Rightarrow x^{2} - 6 x = - 4 y - 17$
$\Rightarrow x^{2} - 6 x + 9 = - 4 y + 8$
$\Rightarrow (x - 3)^{2} = - 4 (y - 2)$
Let $X = x - 3$ and $Y = y - 2$ , then we get
$X^{2} = - 4 Y$
Comparing the above equation with the standard equation of the parabola, we get that the axis of the parabola is given by $X = 0$
Hence, axis of the given parabola will be $x - 3 = 0 \Rightarrow x = 3$
Option $C$ is correct.

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