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Question

The equation of the parabola whose focus is (4,3) and vertex is (4,1)

A
x28x+8y+24=0
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B
x28x8y+8=0
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C
x28x+8y+8=0
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D
x28x+8y+32=0
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Solution

The correct option is A x28x+8y+24=0
x coordinates of the vertex and the focus both are same
axis of the required parabola will be x=4
If we shift the origin at (4,1), then the parabola will be the form of X2=4aY
and a=distance between focus and vertex=2
So required parabola will be
(x4)2=8(y+1)
x28x+8y+24=0

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