The equation of the hyperbola that has a center at (0,0) , a focus at (5 , 0) , and a vertex at (-4 , 0 ) , is \frac(x^2)(A^2)-\frac(y^2)(B^

Thanks for your excellent answer Heureka.

I just want to practice this for myself because I keep forgetting it.   

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The equation of the hyperbola that has a center at (0,0) , a focus at (5 , 0) , and a vertex at (-4 , 0 ) , is

$$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$$

Vertex =( 0 $$\pm$$ A, 0)    So A= 4

     $$\\Focus=(0\pm \sqrt{A^2+B^2},0)\\\\
\sqrt{16+B^2}=5\\
4^2+B^2=5^2\\
B=3$$

$$\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$$

The equation of the hyperbola that has a center at (0,0) , a focus at (5 , 0) , and a vertex at (-4 , 0 ) , is (x^2)/(A^2)-(y^2)/(B^2)=1

$$\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$$

vertex at (-4 , 0 ):

$$\small{\text{$ \begin{array}{l} (-4,0) =(\pm A ,0)\\ A = \pm4 \end{array} $}}$$

focus at (5 , 0) :

$$\small{\text{$ \begin{array}{l} (5,0) =(\pm \sqrt{A^2+B^2} ,0)\end{array} $}}\\ \begin{array}{rcl} A^2+B^2 &=& 5^2 = 25\\
(\pm4)^2+B^2 &=& 25\\
B^2 &=& 25-16 = 9\\ B &=& \pm 3
\end{array} $}}$$

A = $$\small{\text{$\pm 4$}}$$

B = $$\small{\text{$\pm 3$}}$$

$$\dfrac{x^2}{(\pm 4)^2}-\dfrac{y^2}{(\pm 3)^2}=1$$