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# 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

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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
+19207
+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
+19207
+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
+92221
+5

Melody  Jun 16, 2015
#3
+92221
+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
+85805
0

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

CPhill  Jun 16, 2015
#5
+92221
0

Thanks Chris

Melody  Jun 16, 2015

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