The required foci are (±√22, 0) and asymptote are y = ±4/√6x.
Equation of the hyperbola is given i.e. x^2/6 -y^2/16=1.
Foci and asymptote to be determine.
What is hyperbola?
Hyperbola is the cross-section from the cone.
[tex]x^2/6 -y^2/16=1[/tex]
From above equation a =√6 and b = 4 and c =[tex]\sqrt{6+16}[/tex] or √22 can be evaluated.
Now,
Asymptote = (±b(x-h)/a, 0)
y = (±4x/√6, 0)
Foci = (h+c, k )
Here, h and k = 0
Foci = (c, 0 )
Foci = (±√22, 0 )
Thus, the required foci are (±√22, 0) and asymptote are y = ±4/√6x.
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