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Question

The equation of straight line passing through the point (3,6) and cutting y=x orthogonally is

A
4x+y=18
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B
x+y=9
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C
4xy=6
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D
xy=9
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Solution

The correct option is A 4x+y=18
y=xx,y>0
Now squaring both the sides we get
y2=x
Any line orthogonal to parabolic curve is normal.
So, equation of normal is y=tx+at3+2at
Here, a=14
Normal passes through (3,6)
6=3t+t34+2t424+12t=t3+2tt310t24=0
By observation, t=4 is one solution,
(t4)(t2+4t+6)=0

So, t=4 is the only real solution
The required equation of normal is
y=4x+644+84y+4x=18

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