Given:
The equation of the hyperbola is given as,
[tex]\frac{y^2}{25}-\frac{x^2}{4}=1........(1)^{}[/tex]
The objective is to graph the equation of the hyperbola.
Explanation:
The general equation of hyperbola open in the vertical axis of up and down is,
[tex]\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\text{ . . . . . . . .(2)}[/tex]
Here, (h,k) represents the center of the hyperbola.
The focal length can be calculated as,
[tex]c=\sqrt[]{a^2+b^2}\text{ . . . . (3)}[/tex]
On plugging the values of a and b in equation (3),
[tex]\begin{gathered} c=\sqrt[]{5^2+2^2} \\ =\sqrt[]{25+4} \\ =\sqrt[]{29} \end{gathered}[/tex]
The foci can be calculated as,
[tex]\begin{gathered} F(h,k\pm c)=F(0,0\pm\sqrt[]{29}) \\ =F(0,\pm\sqrt[]{29}) \end{gathered}[/tex]
The vertices can be calculated as,
[tex]\begin{gathered} V(h,k\pm a)=V(0,0\pm5) \\ =V(0,\pm5) \end{gathered}[/tex]
To obtain graph:
The graph of the given hyperbola can be obtained as,
Hence, the graph of the given hyperbola is obtained.
