Question

A variable line, drawn through the point of intersection of the straight lines x3+y2=1 & x2+y3=1, meets the coordinate axes in A & B. The locus of the mid point of AB is the curve

A
6(x+y)=10xy
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B
(x+y)=10xy
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C
xy=6(x+y)
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D
2(x+y)=6xy
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Solution

The correct option is C 6(x+y)=10xyThe equation of line through the point os intersection of x3+y2−1=0 and x2+y3−1=0 is of the form(x3+y2−1)+λ(x2+y3−1)=0Putting y=0 we get point A as {6(1+λ)2+3λ,0}Putting x=0 we get point B as {0,6(1+λ)3+2λ}If (h,k) be the mid-point of AB, then 2h=6(1+λ)2+3λ and 2k=6(1+λ)3+2λIn order to find the locus of (h,k) we have to eliminate the variable λNow 12h+12k=(2+3λ)+(3+2λ)6(1+λ)=5(1+λ)6(1+λ)⇒h+k2hk=56⇒6(h+k)=10hkHence, the locus of mid-point (h,k) is 6(x+y)=10xy.

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