answer:The equation has the form y2a2−x2b2=1 y 2 a 2 − x 2 b 2 = 1 , so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set x=0 x = 0 , and solve for y y .
The Equation of the Hyperbola with vertices [tex](0, \pm\sqrt{54} )[/tex] and foci at [tex](0, \pm\sqrt{89} )[/tex] is [tex]\frac{y^{2}}{54} - \frac{x^{2}}{35} =1[/tex]
Equation of the Hyperbola [tex]\frac{y^{2}}{54} - \frac{x^{2}}{35} =1[/tex]
There are two standard equations of the Hyperbola. These equations are based on the transverse axis and the conjugate axis of each of the hyperbola. The standard equation of the hyperbola is [tex]\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} =1[/tex]
has the transverse axis as the x-axis and the conjugate axis is the y-axis. Further, another standard equation of the hyperbola is [tex]\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} =1[/tex].
Given vertices are:[tex](0, \pm\sqrt{54} )[/tex] and foci at [tex](0, \pm\sqrt{89} )[/tex]
The standard form of equation of Hyperbola is:
[tex]\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} =1[/tex]
We have , a= [tex]\sqrt{54}[/tex]
So, [tex]a^{2}=54[/tex]
Focus= [tex](0, \pm\sqrt{89} )[/tex]
c= [tex]\sqrt{89}[/tex] and [tex]c^{2}=89[/tex]
Now, b= [tex]\sqrt{c^{2}-a^{2}}[/tex]
=[tex]\sqrt{89-54}[/tex]
=[tex]\sqrt{35}[/tex]
So, [tex]\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} =1[/tex]
[tex]\frac{y^{2}}{54} - \frac{x^{2}}{35} =1[/tex]
Hence, Equation of the Hyperbola [tex]\frac{y^{2}}{54} - \frac{x^{2}}{35} =1[/tex]
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