# 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 ${x}^{2}+2{y}^{2}-2x-4y=-1$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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