Given the following equation of the hyperbola:
![25x^2-16y^2=400](https://img.qammunity.org/2023/formulas/mathematics/college/zzo74wv1e4sqn593j18mvzswdkgyk8u8zn.png)
The standard form of the hyperbola will be as follows:
![((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1](https://img.qammunity.org/2023/formulas/mathematics/college/p9z9gumo8hk2p84uo05au4dcim97672l9q.png)
We will rewrite the given equation to be like the standard form:
![\begin{gathered} 25x^2-16y^2=400\rightarrow(/400) \\ (25x^2)/(400)-(16y^2)/(400)=(400)/(400) \\ \\ (x^2)/(16)-(y^2)/(25)=1 \\ \\ (x^2)/(4^2)-(y^2)/(5^2)=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/304p1i5tz5r72f8unkoqvc83gs7uvxh74v.png)
Compare the last equation with the standard form:
So, the answer will be as follows:
The value of h = 0
The value of k = 0
The value of a = 4
The value of b = 5