Question

A variable straight line is drawn through the point of intersection of the straight lines x2+y3=1 and x3+y2=1 and meets the coordinate axes at A and B. Then the locus of the mid-point of AB is

A
3x+3y5xy=0
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B
3x3y5xy=0
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C
5x+5y3xy=0
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D
5x5y3xy=0
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Solution

The correct option is A 3x+3y−5xy=0Given, x2+y3=1 ...(1) x3+y2=1 ...(2) Solving (1) and (2), we get x=65 and y=65 ∴ Intersection point P is (65,65) Now, variable line passing through P meets the axes at points A and B. Let the mid-point of AB be M(h,k) whose locus is to be found. Then coordinates of A and B are (2h,0) and (0,2k) respectively. Clearly, points A,B and P are collinear. Then Δ=∣∣ ∣ ∣ ∣∣2h0102k165651∣∣ ∣ ∣ ∣∣=0 ⇒2h(2k−65)+1(0−2k⋅65)=0 ⇒4hk−125h−125k=0 ⇒3h+3k−5hk=0 Hence, the locus of M is 3x+3y−5xy=0

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