Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3 divided by 2.x

Answer 1

The standar form for the hyperbola is:

 (y^2/a^2)-(x^2/b^2)=1

 a=6
 a^2=36

 The equation of the asymptotes is:

 y=±(a/b)x

 a/b=3/2
 b=4
 b^2=16

 Therefore, the equation is:

 (y^2/a^2)-(x^2/b^2)=1
 (y^2/36)-(x^2/16)=1

 

Answer 2

Answer:

The equation of hyperbola with vertices at (0,±6) and asypmtotes at $y= ±\frac{3}{2} x$

Step-by-step explanation:

Given

vertices of hyperbola at (0,±6)

It means hyperbola along y axis and centre at (0,0)

Asymptotes $y=±\frac{3}{2}$

We know that equation of hyperbola along y axis

 $\frac{y^2}{a^2} -\frac{x^2}{b^2} =1$

a=6

We know that the general equation of hyperbola asymptotes and vertcal transverse axis

    y=$±\frac{a}{b} x$

 $\frac{3}{2} x$= $\frac{6}{b} x$

Both side cancel x we get

$b=\frac{6\times2}{3}$

b=4

When we take $-\frac{3}{2} =-\frac{6[tex]b=\frac{-6\times (-2)}{3}$}{b}x[/tex]

Then we get b=4

Now, put the value of a and b in the equation of hyperbola we get

$\frac{y^2}{6^2} -\frac{x^2}{4^2}$=1

$\frac{y^2}{36} -\frac{x^2}{16} =1$

Hence, the required standard equation of hyperbola

$\frac{y^2}{36} -\frac{x^2}{16} =1$.