The vertex of the parabola $x^{2} + y^{2} - 2 x y - 4 x - 4 y + 4 = 0$ is at?

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Solution

For a parabola, $\frac{d y}{d x} = 0$ at the vertex. Hence,
$2 x + 2 y \frac{d y}{d x} - 2 x \frac{d y}{d x} - 2 y - 4 - 4 \frac{d y}{d x} = 0$
$2 x - 2 y - 4 = 0$
$x = 2 + y$
Putting it back in the equation, we get
$(2 + y)^{2} + y^{2} - 2 y (y + 2) - 4 (y + 2) - 4 y + 4 = 0$
$4 + 4 y + y^{2} + y^{2} - 2 y^{2} - 4 y - 4 y - 8 - 4 y + 4 = 0$
$- 8 y = 0$
$y = 0$
$x = 2$

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