The conic represented by the equation $x^{2} + y^{2} + 2 x y + 4 x - 6 y + 17 = 0$ is

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Solution

$x^{2} + y^{2} + 2 x y + 4 x - 6 y + 17 = 0$
On comparing the given equation with $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , we get
$a = b = 1, c = 17, g = 2, f = - 3, h = 1$
$∴ Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = - 25 \neq 0$
and $h^{2} - a b = 0$
Hence, $Δ \neq 0$ and $h^{2} = a b$
So the given equation represents a parabola.

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