Question

Find the eccentricity of the hyperbola.
x^2 - 2y^2 = 6

Answer 1

can you tell what the "a" and "b" components are?

keep in mind that a hyperbola's eccentricity is the same as in an ellipse, that is \(\bf e=\cfrac{c}{a}\)

Answer 2

divide by 6
write in the form
\[\frac{ x^2 }{ a^2 }-\frac{ y^2 }{ b^2 }=1\]

Answer 3

then \[b^2=a^2\left( e^2-1 \right)\]

Answer 4

\(\bf \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1
\\ \quad \\ \quad \\
x^2-2y^2=6\implies \cfrac{x^2}{6}-\cfrac{\cancel{ 2 }y^2}{\cancel{ 6 }}=1\implies \cfrac{x^2}{6}-\cfrac{y^2}{3}=1
\\ \quad \\
\cfrac{(x-0)^2}{6}-\cfrac{(y-0)^2}{3}=1\implies \cfrac{(x-0)^2}{(\sqrt{6})^2}-\cfrac{(y-0)^2}{(\sqrt{3})^2}=1\)

as surjithayer suggested

Answer 5

well.. keep in mind for a hyperbola, \(\bf c=\sqrt{a^2+b^2}\)

so \(\bf a^2=6\qquad b^2=3\qquad c=\sqrt{a^2+b^2}\to \sqrt{6+3}\to \sqrt{9}\to 3
\\ \quad \\
thus\qquad e=\cfrac{c}{a}\implies e=\cfrac{3}{\sqrt{6}}\)

Answer 6

and I gather you'd need to rationalize that, by mulplying top and bottom by \(\bf \sqrt{6}\)

Answer 7

just to clarify \(\bf \cfrac{ x^2 }{ 6 } - \cfrac{ y^2 }{ 3} = 1\implies \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1
\\ \quad \\
meaning\implies a^2=6\implies a=\sqrt{6}\qquad b^2=3\implies b=\sqrt{3}\)

Answer 8

\[\frac{ b^2 }{ a^2 }=e^2-1,e^2=\frac{ 3 }{ 6 }+1=\frac{ 3 }{ 2 },e=\sqrt{\frac{ 3 }{ 2 }}\]