Find the equation of the hyperbola whose foci are (8, 3) and (0, 3) and eccentricity is 4/3.

Options

A. `49x^(2)-3y^(2)=147`
B. `7x^(2)-9y^(2)=63`
C. `7(x-4)^(2)-9(y-3)^(2)=63`
D. `7(x+4)^(2)-9(y+3)^(2)=63`

Answer

C

Explanation

The centre of the hyperbola is the mid-point 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 length 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`....`(i)`
`:.` Distance between the two foci `=2ae`
`sqrt((8-0)^(2)+(3-3)^(2))=2ae`
`ae=4impliesa=3` `[ :' e=4//3]`
`:.b^(2)=a^(2)(e^(2)-1)impliesb^(2)=9((16)/(9)-1)=7`
Substituting the value of `a` and `b` in `(i)`, we find that the equation of the hyperbola is
`((x-4)^(2))/(9)-((y-3)^(2))/(7)=1` or, `7(x-4)^(2)-9(y-3)^(2)=63`