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

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
asked 2020-11-12
Find and calculate the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections \(x^2 + 2y^2 - 2x - 4y = -1\)

Answers (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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