Which is the equation of a hyperbola centered at the origin with y-intercepts +12 -12, and asymptote y=3x/2



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Respuesta :

Answer:

[tex]\frac{y^2}{144}-\frac{x^2}{64}=1[/tex]

Step-by-step explanation:

The equation of a hyperbola centered at the origin with vertices on the y-axis is given by: [tex]\frac{y^2}{a^2}-\frac{x^2}{b^2}=1[/tex]

The vertices of the hyperbola are the y-intercepts (0,12) and (0,-12)

This implies that:

[tex]2a=|12--12|[/tex]

[tex]2a=24[/tex]

[tex]a=12[/tex]

The asymptote equation of a hyperbola is given by:

[tex]y=\pm\frac{a}{b}x[/tex]

The given hyperbola has asymptote: [tex]y=\pm\frac{3}{2} x[/tex]

By comparison; [tex]\frac{a}{b}=\frac{3}{2}[/tex]

[tex]\implies \frac{12}{b}=\frac{12}{8}[/tex]

[tex]\implies b=8[/tex]

The required equation is:

[tex]\frac{y^2}{12^2}-\frac{x^2}{8^2}=1[/tex]

Or

[tex]\frac{y^2}{144}-\frac{x^2}{64}=1[/tex]

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Answer: Answer D

\frac{y^2}{144}-\frac{x^2}{64}=1