# Find and calculate the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections x^2 + 2y^2 - 2x - 4y = -1

Question
Conic sections
Find and calculate the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections $$x^2 + 2y^2 - 2x - 4y = -1$$

2020-11-13
Given $$x^2 + 2y^2 - 2x - 4y = -1$$
$$(x^2 - 2x) + 2 (y^2 - 2y) = -1$$
$$(x^2 - 2x + 1) + 2 (y^2 - 2y + 1) = -1 + 1 + 2$$
$$(x - 1)^2 + 2(y - 1)^2 = 2$$
$$\frac{(x - 1)^2}{2} + \frac{2(y - 1)^2}{2} = \frac{2}{2}$$
$$\frac{(x - 1)^2}{2} + (y - 1)^2 = 1$$ It is in the form of $$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$
$$a^2 = 2 , b^2 = 1$$
$$a = \sqrt2 , b = 1$$ Distance between center and focus $$c = \sqrt{2 - 1}$$
$$= 1$$ Hence Center$$(h, k) = (1, 1)$$ Foci $$= (h \pm c, k)$$
$$=(1 \pm 1, 1)$$ Vertices $$= (h \pm a, k)$$
$$=(1 \pm \sqrt2, 1)$$

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