Find the equation of the hyperbola whose foci are `(8,3)a n d(0,3)` and eccentricity `=4/3dot`

Correct Answer - `((x-4)^(2))/(9)-((y-3)^(2))/(7)=1`
The centre of the hyperbola is the midpoint of the line joining the two foci.
So, the coordinates of the centre are `((8+0)//2,(3+3)//2),i.e., (4,3).`
Let 2a and 2b be the lengths of transverse and conjugate axes and let e be the eccentricity.
Then the equation of the hyperbola is
`((x-4)^(2))/(a^(2))-((y-3)^(2))/(b^(2))=1" (1)"`
Now, the distance between the foci is 2ae or ae = 4
`"or a = 3 "(becausee=(4)/(3))`
Now, `b^(2)=a^(2)(e^(2)-1)=9(-1+(16)/(9))=7`
Thus, the equation of the hyperbola is
`((x-4)^(2))/(9)-((y-3)^(2))/(7)=1`