Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

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

This implies that:

$2a=|12--12|$

$2a=24$

$a=12$

The asymptote equation of a hyperbola is given by:

$y=\pm\frac{a}{b}x$

The given hyperbola has asymptote: $y=\pm\frac{3}{2} x$

By comparison; $\frac{a}{b}=\frac{3}{2}$

$\implies \frac{12}{b}=\frac{12}{8}$

$\implies b=8$

The required equation is:

$\frac{y^2}{12^2}-\frac{x^2}{8^2}=1$

Or

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

Answer 2

Answer: Answer D

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