# Write the equation of a hyperbola with a center at (-5, -3), vertices at (-5, -5) and (-5, -1) and co-vertices at (-11, -3) and (1, -3)

*print*Print*list*Cite

Student Comments

aruv | Student

Lenth of of transverse axis=`sqrt((-5+5)^2+(-5+1)^2)=sqrt(16)`

`=4`

so semi transverse axis=2

Length of conjugate axis=`sqrt((-11-1)^2+(-3+3)^2)=sqrt(144)`

`=12`

So semi conjugate axis=6

Thus equation of the hyperbola

`(x-(-5))^2/2^2-(y-(-3))^2/6^2=1`

`(x+5)^2/4-(y+3)^2/36=1`