Correct Answer - 3
3 Given that equation of tangent at vertex is 3x+4y+k=0.
Let the equation of axis be `4x-3y+lamda=0`
Therefore, equation of parabola is
`(4x-3y+lamda)^(2)=mu(3x+4y+k)` (1)
Given equation of parabola is
`16x^(2)-24xy+9y^(2)+14x+2y+7=0` (2)
Comparing equations (1) and (2), we get
`8lamda-3mu=14,-6lamda-4mu=2,lamda^(2)-kmu=7`
Solving, we get
`lamda=1,mu=-2andk=3`.