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Prove that product of parameters of four concyclic points on the hyperbola xy=c^(2) is 1. Also, prove that the mean of these four concyclic points bisects the distance between the centres of the hyperbola and the circle. |
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Answer» Solution :Given equation of hyperbola is `xy=C^(2)` Let four conyclic point on the hyperbola be `(x_(i),y_(i)),i=1,2,3,4`. Let the equation of the circle through point A, B, C and D be `x^(2)+y^(2)+2gx+2fy+d=0"(1)"` Solving circle and hyperbola, we get `x^(2)+(c^(4))/(x^(2))+2gx+2f*(c^(2))/(x)+d=0` `rArr""x^(4)+2gx^(3)+dx^(2)+2fc^(2)x+c^(4)=0"(2)"` `therefore"Product of roots,"x_(1)x_(2)x_(3)x_(4)=c^(4)` Now, `(x_(i),y_(i))-=(ct_(i),(c)/(t_(i)))` `therefore""(ct_(1))(ct_(2))(ct_(3))(ct_(4))=c^(4)` `rArr""t_(1)t_(2)t_(3)t_(4)=1` Now, MEAN of points `(x_(i),y_(i)),i=1,2,3,4," is "((sum_(i=1)^(4)x_(i))/(4),(sum_(i=1)^(4)y_(i))/(4))`. From Eq. (2), `x_(1)+x_(2)+x_(3)+x_(4)=-2G` `rArr""(sum_(i=1)^(4)x_(i))/(4)=-(g)/(2)` Also, `sum_(i=1)^(4)y_(i)=sum_(i=1)^(4)(c^(2))/(x_(i))=(c^(2)sumx_(1)x_(2)x_(3))/(x_(1)x_(2)x_(3)x_(4))=(c^(2))/(c^(4))(-2fc^(2))=-2f` `therefore""(sumy_(i))/(4)=-(f)/(2)` Thus, `((sum_(i=1)^(4)x_(i))/(4),(sum_(i=1)^(4)y_(i))/(4))-=(-(g)/(2),-(f)/(2))`. |
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