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Normals are drawn from the external point `(h,k)` to the rectangular hyperbola `xy=c^(2)`. If circle are drawn through the feet of these normals taken three at a time then centre of circle lies on another hyperbola whose centre and eccentricity isA. `(h/2, k/2)`B. `(h,k)`C. `sqrt(2)`D. `sqrt(2)+1` |
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Answer» Correct Answer - A::C Equation of normal to hyperbola `xy=c^(2)` at `(ct, c/t)` which passes through `(h,k)` is `ct^(4)-ht^(3)+kt-c=0` Roots of this equation is `t_(1),t_(2),t_(3)` and `t_(4)` Let equation of circle be `x^(2)+y^(2)-2gx-2fy+p=0` If `(ct, c/t)` lies on this `c^(2)t^(4)-2gct^(3)+pt^(2)-2fct+c^(2)=0` It roots are `t_(1), t_(2), t_(3)` and `t_(4)` We get `t_(4)=-t_(4)` also `c/(t_(4))-c/(t_(4))=k-2f` So locus of centre is `4c^(2)=(h-2g)(k-2f)` Centre lies on hyperbola `(x-h/2)(y-k/2)=c^(2)` `c(h/2,k/2)` and `c=sqrt(2)` |
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