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From a point, P perpendicular PM and PN are drawn to x and y axes, respectively. If MN passes through fixed point (a,b), then locus of P is

A

`xy=ax+ by `

B

`xy=ab`

C

`xy=bx+ay`

D

`x+y=xy`

Text Solution

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The correct Answer is:
C

Let P(h,k)
So, points M and N are (h,0) and (0,k), respectively. `MN` passes through the point `Q(a,b)`
So, M,N and Q are collinear.
`therefore` Slope of MN=Slope of QM
`therefore (k)/(h)=(b-0)/(a-h)`
`therefore ak-hk=-bh`
So locus is `bx+ay=xy`.
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