Answer:
The equation of hyperbola with vertices at (0,±6) and asypmtotes at [tex]y= ±\frac{3}{2} x[/tex]
Step-by-step explanation:
Given
vertices of hyperbola at (0,±6)
It means hyperbola along y axis and centre at (0,0)
Asymptotes [tex]y=±\frac{3}{2}[/tex]
We know that equation of hyperbola along y axis
[tex]\frac{y^2}{a^2} -\frac{x^2}{b^2} =1[/tex]
a=6
We know that the general equation of hyperbola asymptotes and vertcal transverse axis
y=[tex] ±\frac{a}{b} x[/tex]
[tex]\frac{3}{2} x[/tex]= [tex]\frac{6}{b} x[/tex]
Both side cancel x we get
[tex]b=\frac{6\times2}{3}[/tex]
b=4
When we take [tex]-\frac{3}{2} =-\frac{6[tex]
b=\frac{-6\times (-2)}{3}[/tex]}{b}x[/tex]
Then we get b=4
Now, put the value of a and b in the equation of hyperbola we get
[tex]\frac{y^2}{6^2} -\frac{x^2}{4^2}[/tex]=1
[tex]\frac{y^2}{36} -\frac{x^2}{16} =1[/tex]
Hence, the required standard equation of hyperbola
[tex]\frac{y^2}{36} -\frac{x^2}{16} =1[/tex].
