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Q.

Let Px0,y0 be the point on the hyperbola 3x24y2=36, which is nearest to the line 3x+2y=1, Then 2y0x0 is equal to 

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a

3

b

-3

c

9

d

-9

answer is D.

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Detailed Solution

detailed_solution_thumbnail

Given hyperbola 3x24y236=0
Diff. w.r.to x
6x8ydydx=0dydx=3x4y
Given dydxatpx0,y0=32
3x04y0=32x0=2y0
Since,x0,y0 lie on 3x24y236=0                  3x024y0236=0
34y024y02=36y0=±32
If y0=32;x0=32  point A=32,32
Perpendicular distance from A32,32 to 3x+2y=1 is |92+321|13=62+113
If y0=32x0=32  point  B=32,32
Perpendicular distance from

 B32,-32 to 3x+2y1=0 is  |92321|13=|621|13
Point B is the nearest to 3x+2y1=0
2y0x0=2y0+2y0=32y0=9

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