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
where A =_____
B =_____
Thanks for your excellent answer Heureka.
I just want to practice this for myself because I keep forgetting it.
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 = $$\small{\text{$\pm 3$}}$$
$$\dfrac{x^2}{(\pm 4)^2}-\dfrac{y^2}{(\pm 3)^2}=1$$
Thanks for your excellent answer Heureka.
I just want to practice this for myself because I keep forgetting it.
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
$$\\Focus=(0\pm \sqrt{A^2+B^2},0)\\\\
\sqrt{16+B^2}=5\\
4^2+B^2=5^2\\
B=3$$
$$\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$$