Question

The equation of the hyperbola that has a center at ( 6 , 9 ) , a focus at ( 11 , 9 ) , and a vertex at ( 9 , 9 ) (x−C)^2/A^2−(y−D)^2/B^2=1

Answer 1

Answer:

$A=-3$

$C=6$

$B=4$

$D=9$

Step-by-step explanation:

From the question we are told that:

Center of hyperbola at ( 6 , 9 )

Focus of hyperbola at ( 11 , 9 )

Vertex of hyperbola at ( 9 , 9)

Equation of hyperbola $\frac{(x-C)^2}{A^2} -\frac{(y-D)^2}{B^2}=1$

Generally the C and D of the hyperbola equation is mathematically given by

$Centers (6,9)$

$C=6$

$D=9$

Generally the A and B a of the hyperbola equation is mathematically given by

 $A=x_c-x_v$

 $A=6-9$

 $A=-3$

 $C'=x_c-x_f$

 $C'=6-11$

 $C'=-5$

Therefore with Center,Focus ,Vertex on the same line

 $B^2=C'^2-A^2$

 $B^2=(-5^2)-(-3^2)^2$

 $B^2=(25)-(9)$

 $B^2=16$

 $B=4$