If the point `(2a,a)` lies inside the parabola `x^(2) -2x - 4y +3 = 0`, then a lies in the interval

Options

A. `[(1)/(2),(3)/(2)]`
B. `((1)/(2),(3)/(2))`
C. `(1,3)`
D. `((-3)/(2),(-1)/(2))`

Answer

B

Explanation

The parabola is `(x-1)^(2) =4 (y-(1)/(2))` and origin lies outside the parabolic region, (0,0) makes `x^(2) - 2x - 4y +3` positive.
`:. (2a,a)` should make `x^(2) -2x -4y +3` negative i.e., `4a^(2) - 8a + 3 lt 0`
`rArr (2a-1) (2a-3) lt 0`
Thus, a belongs to the open interval `((1)/(2),(3)/(2))`.