Ramsey

2020-11-12

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

berggansS

Skilled2020-11-13Added 91 answers

Given ${x}^{2}+2{y}^{2}-2x-4y=-1$

$({x}^{2}-2x)+2({y}^{2}-2y)=-1$

$({x}^{2}-2x+1)+2({y}^{2}-2y+1)=-1+1+2$

$(x-1{)}^{2}+2(y-1{)}^{2}=2$

$\frac{(x-1{)}^{2}}{2}+\frac{2(y-1{)}^{2}}{2}=\frac{2}{2}$

$\frac{(x-1{)}^{2}}{2}+(y-1{)}^{2}=1$
It is in the form of
$\frac{(x-h{)}^{2}}{{a}^{2}}+\frac{(y-k{)}^{2}}{{b}^{2}}=1$

${a}^{2}=2,{b}^{2}=1$

$a=\sqrt{2},b=1$
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 $=(h\pm a,k)$

$=(1\pm \sqrt{2},1)$

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