The equation of the parabola whose focus is (4,−3) and vertex is (4,−1)
A
x2−8x+8y+24=0
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B
x2−8x−8y+8=0
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C
x2−8x+8y+8=0
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D
x2−8x+8y+32=0
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Solution
The correct option is Ax2−8x+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 (x−4)2=−8(y+1) ⇒x2−8x+8y+24=0