Given the equation of the hyperbola
[tex]16x^2-4y^2=64[/tex]The standard equation of hyperbola is of the form
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]Write the given equation of hyperbola in standard form.
Divide the given equation by 64.
[tex]\begin{gathered} \frac{16x^2}{64}-\frac{4y^2}{64}=1 \\ \frac{x^2}{4}-\frac{y^2}{16}=1 \end{gathered}[/tex]Comparing with the standard form gives
[tex]a=2,b=4[/tex]Vertices are
[tex]\begin{gathered} =(\pm a,0) \\ =(\pm2,0) \end{gathered}[/tex]Find the eccentricity.
[tex]\begin{gathered} e=\frac{\sqrt[]{a^2+b^2}}{a} \\ =\frac{\sqrt[]{4+16}}{2} \\ =\frac{\sqrt[]{20}}{2} \\ =\sqrt[]{5} \end{gathered}[/tex]Foci are
[tex]\begin{gathered} =(\pm ae,0) \\ =(\pm2\sqrt[]{5},0) \end{gathered}[/tex]Asymptotes are bx-ay = 0 and bx+ay=0
[tex]\begin{gathered} 4x-2y=0\text{ and 4x+2y=0} \\ 2x-y=0\text{ and 2x+y=0} \end{gathered}[/tex]
