Answer:
[tex]\dfrac{y^2}{64}-\dfrac{x^2}{9}=1[/tex] .
Step-by-step explanation:
Since vertices lie on y-axis. So, it is a vertical parabola of the form
[tex]\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1[/tex]
where, (h,k) is center, [tex](h,k\pm c)[/tex] is focus and [tex](h,k\pm a)[/tex] is vertex.
Center is (0,0). So, h=0 and k=0.
Foci are [tex](0,\pm \sqrt{73})[/tex]. So [tex]c=\sqrt{73}[/tex].
Vertices are [tex](0,\pm 8)[/tex]. So [tex]a=8[/tex].
We know that,
[tex]a^2+b^2=c^2[/tex]
[tex]8^2+b^2=(\sqrt{73})^2[/tex]
[tex]b^2=73-64[/tex]
[tex]b=3[/tex]
Put h=0,k=0, a=8 and b=3 in equation (1).
[tex]\dfrac{(y-0)^2}{8^2}-\dfrac{(x-0)^2}{3^2}=1[/tex]
[tex]\dfrac{y^2}{64}-\dfrac{x^2}{9}=1[/tex]
Therefore, the required equation is [tex]\dfrac{y^2}{64}-\dfrac{x^2}{9}=1[/tex] .
