Answer:
[tex]\frac{y^2}{36}-\frac{x^2}{196}=1[/tex]
Step-by-step explanation:
From the question we are told that:
Vertices of hyperbola at [tex](0,\pm6)[/tex]
Asymptotes at [tex]Y= \pm 3/7x[/tex]
Generally the expression for Vertices of hyperbola is mathematically given by
[tex](0,\pm a)[/tex]
Therefore
[tex](0,\pm a)=(0,\pm6)[/tex]
Comparing variables
[tex]a=6[/tex]
Generally the expression for Asymptotes of hyperbola is mathematically given by
[tex]y=\pm \frac{a}{b}x[/tex]
Therefore
[tex]\pm \frac{a}{b}x= \pm \frac{3}{7}x[/tex]
Therefore
[tex]\frac{6}{b}= \frac{3}{7}[/tex]
[tex]b=\frac{42}{3}[/tex]
[tex]b=14[/tex]
Generally the equation for Hyperbola of hyperbola is mathematically given by
[tex]\frac{y^2}{a^2}-\frac{x^2}{b^2}=1[/tex]
[tex]\frac{y^2}{6^2}-\frac{x^2}{14^2}=1[/tex]
[tex]\frac{y^2}{36}-\frac{x^2}{196}=1[/tex]
