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Question

lf two conics whose equations refer to rect- angular axes are ax2+2hxy+by2+2gx+2fy+c=0
and a′x2+2h′xy+b′y2+2g′x+2fy+c′=0 intersect in four concyclic points, then

A
h(ab)=h(ab)
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B
h(ab)=h(ab)
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C
h(ab)=h(ab)
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D
h(ab)=h(ab)
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Solution

The correct option is A h(ab)=h(ab)
S1:ax2+2hxy+by2+2gx+2fy+c=0
S2:ax2+2hxy+by2+2gx+2fy+c=0
four points intersects on the curve: S1+kS2=0
ax2+2hxy+by2+2gx+2fy+c+k(ax2+2hxy+by2+2gx+2fy+c)=0
Since, they are con-cyclic points
Therefore, coefficient of x2= coefficient of y2 and coefficient of xy=0
a+ka=b+kb
2h+2kh=0
Eliminating k, we get h(ab)=h(ab)

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