A) \[{{y}^{2}}+4y+4x-8=0\]
B) \[{{y}^{2}}+4y-4x+8=0\]
C) \[{{y}^{2}}+4y-4x-8=0\]
D) None of these
Correct Answer: C
Solution :
[c] Since the axis is horizontal an vertex is \[(-3,-2),\] \[\therefore \] The equation of the parabola must be of the form \[{{(y+2)}^{2}}=4a(x+3)\] It passes through \[(1,2),\] so \[16=16a\].i.e., a = 1. Hence, the equation of the required parabola is \[{{(y+2)}^{2}}=4(x+3)\] or \[{{y}^{2}}+4y-4x-8=0\]You need to login to perform this action.
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