If the focus of the parabola `x^2-k y+3=0` is (0,2), then a values of `k` is (are) 4 (b) 6 (c) 3 (d) 2

Options

A. 4
B. 6
C. 3
D. 2

Answer

B, D

Explanation

2,4 The given parabola is
`x^(2)-ky+3=0`
`orx^(2)=k(y-(3)/(k))`
Let `x=Y,y-(3)/(k)=X`
Then the parabola is
`Y^(2)=kX`
whose focus is (0,k/4).
Therefore, the focus of
`x^(2)=k(y-(3)/(k))` is
`(0,(3)/(k)+(k)/(4))-=(0,2)`
`:.(3)/(k)+(k)/(4)=2`
`or12+k^(2)=8k`
`ork^(2)-8k+12=0`
`or(k-6)(k-2)=0`
`ork=2,6`