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Question

A(−4,0),B(4,0) are two points. M and N are the variable points of y−axis such that M lies below N and MN=4. Line joining AM and BN intersect at P. Locus of P is

A
2xy16x2=0
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B
2xy+16x2=0
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C
2xy+16+x2=0
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D
2xy16+x2=0
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Solution

The correct option is D 2xy16+x2=0
Let M=(0,h)
Therefore N will be N=(0,h+4)
Therefore, equation of AM by using slope intercept form will be
x4+yh=1
By rearranging the terms we get
Therefore yh=x+44
h=4yx+4 ...(i)
Similarly equation of BN is
x4+yh+4=1
4x4=yh+4
h+4=4y4x
h=4y4x4 ...(ii)
Equating (i) and (ii) we get
4y4x4=4y4+x
4y4x=4y4+x+4
now multiply the above equation with (4 - x).(4 + x)
we get 4y×(4+x)=4y×(4x)+4×(4+x)×(4x)
so we get
x2+2xy16=0

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