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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^

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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 (x^2)/(A^2)-(y^2)/(B^2)=1

where A =_____

B =_____

Guest Jun 16, 2015

#2
+92251
+10

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

http://web2.0calc.com/questions/the-equation-of-the-hyperbola-that-has-a-center-at-1-2-a-focus-at-4-2-and-a-vertex-at-3-2-is-frac-x-c-2-a-2-frac-y#r2

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

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

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

Melody  Jun 16, 2015
Sort:

#1
+19207
+10

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

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

heureka  Jun 16, 2015
#2
+92251
+10

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

http://web2.0calc.com/questions/the-equation-of-the-hyperbola-that-has-a-center-at-1-2-a-focus-at-4-2-and-a-vertex-at-3-2-is-frac-x-c-2-a-2-frac-y#r2

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

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

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

Melody  Jun 16, 2015

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