Let L be a normal to the parabola y2=4x. If L passes through the point (9, 6), then L is given by
A
y−x+3=0
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B
y+3x−33=0
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C
y+x−15=0
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D
y−2x+12=0
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Solution
The correct option is Dy−2x+12=0 Normal to y2=4x, is y=mx−2m−m3 which passes through (9, 6). ⇒6=9m−2m−m3 ⇒m3−7m+6=0 ⇒m=1,2,−3 ∴ Equation of normals are, y−x+3=0,y+3x−33=0 and y−2x+12=0