+0

# 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

0
356
5

The equation of the hyperbola that has a center at (1,2) , a focus at (-4 , 2) , and a vertex at (-3 , 2 ) , is ((x-C)^2)/(A^2)-((y-D)^2)/(B^2)=1

where A =____

B =_____

C =______

D =______

Guest Jun 16, 2015

#1
+18829
+10

The equation of the hyperbola that has a center at (1,2) , a focus at (-4 , 2) , and a vertex at (-3 , 2 ) , is ((x-C)^2)/(A^2)-((y-D)^2)/(B^2)=1

$$\dfrac{(x-C)^2}{A^2}-\dfrac{(y-D)^2}{B^2}=1$$

center at (1,2):

$$\small{\text{ \begin{array}{rcl} x-C &=& 0 \\ 1-C &=& 0 \\ C &=& 1 \\\\ y-D &=& 0 \\ 2-D &=& 0 \\ D &=& 2 \end{array} }}$$

vertex at (-3 , 2 ):

$$\small{\text{ \begin{array}{l} (-3 , 2 ) - (1,2)_{\mathrm{center}} = (-4,0) =(\pm A ,0)\\ A = \pm 4 \end{array} }}$$

focus at (-4 , 2):

$$\small{\text{ \begin{array}{l} (-4 , 2 ) - (1,2)_{\mathrm{center}} = (-5,0) =(\pm \sqrt{A^2+B^2} ,0)\end{array} }}\\ \begin{array}{rcl} A^2+B^2 &=& (-5)^2 = 25\\ (\pm 4)^2+B^2 &=& 25\\ B^2 &=& 25-16 = 9\\ B &=& \pm 3 \end{array} }}$$

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

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

C = 1

D = 2

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

heureka  Jun 16, 2015
Sort:

#1
+18829
+10

The equation of the hyperbola that has a center at (1,2) , a focus at (-4 , 2) , and a vertex at (-3 , 2 ) , is ((x-C)^2)/(A^2)-((y-D)^2)/(B^2)=1

$$\dfrac{(x-C)^2}{A^2}-\dfrac{(y-D)^2}{B^2}=1$$

center at (1,2):

$$\small{\text{ \begin{array}{rcl} x-C &=& 0 \\ 1-C &=& 0 \\ C &=& 1 \\\\ y-D &=& 0 \\ 2-D &=& 0 \\ D &=& 2 \end{array} }}$$

vertex at (-3 , 2 ):

$$\small{\text{ \begin{array}{l} (-3 , 2 ) - (1,2)_{\mathrm{center}} = (-4,0) =(\pm A ,0)\\ A = \pm 4 \end{array} }}$$

focus at (-4 , 2):

$$\small{\text{ \begin{array}{l} (-4 , 2 ) - (1,2)_{\mathrm{center}} = (-5,0) =(\pm \sqrt{A^2+B^2} ,0)\end{array} }}\\ \begin{array}{rcl} A^2+B^2 &=& (-5)^2 = 25\\ (\pm 4)^2+B^2 &=& 25\\ B^2 &=& 25-16 = 9\\ B &=& \pm 3 \end{array} }}$$

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

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

C = 1

D = 2

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

heureka  Jun 16, 2015
#2
+91467
+5

Melody  Jun 16, 2015
#3
+91467
+5

$$\dfrac{(x-C)^2}{A^2}-\dfrac{(y-D)^2}{B^2}=1$$

The equation of the hyperbola that has a center at (1,2) ,

$$\dfrac{(x-1)^2}{A^2}-\dfrac{(y-2)^2}{B^2}=1$$                  and            $$C^2=A^2+B^2$$

a focus at (-4 , 2) ,   and a vertex at (-3 , 2 )

The foci are at   $$(h\pm C,k)$$      so    1-C=-4    C=5

The vertices are at     $$(h\pm A, k)$$      1-A=-3     A=4

$$\\4^2+B^2=5^2\\ B=3$$

$$\\\dfrac{(x-1)^2}{4^2}-\dfrac{(y-2)^2}{3^2}=1\\\\\\ \dfrac{(x-1)^2}{16}-\dfrac{(y-2)^2}{9}=1$$

Melody  Jun 16, 2015
#4
+81029
0

Very nice, heureka and Melody.......

CPhill  Jun 16, 2015
#5
+91467
0

Thanks Chris

Melody  Jun 16, 2015

### 4 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details