We compare this equation with "x^2\/a^2-y^2\/b^2=1"
Eccentricity is "e=\\sqrt{1+b^2\/a^2}=\\sqrt{\\frac{3}{2}}"
The center is "C=(0,0)"
The vertices are "V'=(\u2212a,0)=(\u22122\\sqrt{2},0)" and "V=(a,0)=(2\\sqrt{2},0)"
To find the foci, we need the distance from the center to the foci "c^2=a^2+b^2=12, \\,\\,c=\\pm 2\\sqrt{3}"
The foci are "F'=(-c,0)=(2\\sqrt{3},0)" and "F=(c,0)=(-2\\sqrt{3},0)"
The asymptotes are "x^2\/8-y^2\/4=0,\\,\\, y=\\pm \\frac{1}{\\sqrt{2}}x"
We compare equation of tangent to hyperbola "x_0x\/a^2-y_0y\/b^2=1" with "x+\\lambda y=2"
We have "x-\\frac{a^2}{b^2} \\frac{y_0}{x_0}y=\\frac{a^2}{x_0}, \\,\\, x-2\\frac{y_0}{x_0}y=\\frac{8}{x_0}" and "x_0=4", so "y_0=\\pm2" and "\\lambda=-\\frac{y_0}{2}=\\mp 1"
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