The midpoint of the chord `4x-3y=5` of the hyperbola `2x^2-3y^2 = 12` is

Options

A. `(0, -(5)/(3))`
B. `(2,1)`
C. `((5)/(4),0)`
D. `((11)/(4),2)`

Answer

B

Explanation

Let mid-point of chord of the hyperbola
`(x^(2))/(6) - (y^(2))/(4) = 1` be `(x_(1), y_(1))`.
Therefore, equation of chord is
`T = S_(1)`
`rArr ("xx"_(1))/(6) - (yy_(1))/(4) -1 = (x_(1)^(2))/(6) - (y_(1)^(2))/(4) -1`
`rArr(x_(1))/(6)x-(y_(1))/(4)y=(x_(1)^(2))/(6) -(y_(1)^(2))/(4)`
Comparing it with 4x - 3y = 5, we get
`x_(1) = 2, y_(1) = 1`