Let `y^(2) -5y +3x +k = 0` be a parabola, then
(a) its latus rectum is least when `k = 1`
(b) its latus crectum is independent of k
(c) the line `y = 2x +1` will touch the parabola if `k = (73)/(16)`
(d) `y = (5)/(2)` is the only normal to the parabola whose slope is zero

Options

A. its latus rectum is least when `k = 1`
B. its latus rectum is independent of k
C. the line `y = 2x +1` will touch the parabola if `k = (73)/(16)`
D. `y = (5)/(2)` is the only normal to the parabola whose slope is zero

Answer

B, C, D

Explanation

The equation to the parabola can be written as:
`(y-(5)/(2))^(2) =- 3 (x-(25-4k)/(12))`
The length of the latus rectum is `3.y = 2x+1` is a tangent. So, the quadratic equation `(2x+1)^(2) - 5 (2x+1) +3x +k =0` or `4x^(2) -3x +k-4 =0` must have equal roots. Now roots are equal if `b^(2) = 4ac`
`rArr 9 = 16(k-4) rArr k =(73)/(16)`
`y = (5)/(2)` is normal at the vertex which has slope 0.