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Question

The line x+2y=36 is normal to the parabola x2=12y at the point whose distance from the focus of the parabola is

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Solution

Equation of normal to the parabola x2=4ay at P(2at,at2) in parametric form is x+ty=2at+at3
For parabola x2=12y, we get
x+ty=6t+3t3
Given equation of normal is x+2y=36

Comparing both equations, we get
t=2
Point of contact is P(2×3×2,3×22)=P(12,12)

Focus of the parabola is F(0,3)
PF=122+92=15

Alternate :
x2=12y
2x=12dydx
dydx=x6
Slope of normal at (x1,y1)is 6x1

Now given equation is
x+2y=36
Comparing slope, we get
6x1=12x1=12y1=x2112=12

Therefore, the point P is (12,12)
Focus is F(0,3)
Required distance,
PF=122+92=15

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