The parametric equation of the parabola (x−1)2=−12(y−2) is
A
x=6t,y=2−3t2
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B
x=1+6t,y=2−3t2
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C
x=1+6t,y=2+3t2
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D
x=6t,y=3t2
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Solution
The correct option is Bx=1+6t,y=2−3t2 Here, parabola is of form (x−h)2=−4a(y−k) ∵4a=12⇒a=3,
parametric equation is ⇒y−k=−at2⇒y−2=−3t2 ⇒x−h=2at⇒x−1=6t ⇒x=1+6t,y=2−3t2