[tex]=\text{ }\frac{y^2}{25}-\frac{x^2}{9}[/tex]Explanation:
The standard form of the hyperbola:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}\text{ = }1[/tex]
The center of the hyperbola is at (h, k) = The point where the lines intersect
(h, k) = (0, 0)
For hyperbola:
c² = a² + b²
b is gotten by tracing the value of a to the intersecting line. Trace 3 down to the x axis. you get 5.
[tex]\begin{gathered} a\text{ = distance from center to vertex} \\ \text{the hyperbola coordinate = (0, a) = }(0,\text{ 3)} \\ \text{the other one = (0, -a) = (0, -3)} \\ \text{Hence, a = 3} \end{gathered}[/tex][tex]\begin{gathered} c\text{ = distance from the center to the focus point} \\ b\text{ = 5 and -b = 5} \\ c^2=3^2+5^2 \end{gathered}[/tex][tex]\begin{gathered} c^2\text{ = 9+25 = 34} \\ c\text{ = }\sqrt[]{34} \\ c\text{ = 5.83} \end{gathered}[/tex]
The equation becomes:
[tex]\begin{gathered} \frac{(y-k)^2}{5^2}-\frac{(x-h)^2}{3^2}=\text{ }\frac{(y-0)^2}{5^2}-\frac{(x-0)^2}{3^2} \\ =\text{ }\frac{y^2}{25}-\frac{x^2}{9} \end{gathered}[/tex]
