1.

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 is

Answer»

`(h/2, k/2)`
`(h,k)`
`SQRT(2)`
`sqrt(2)+1`

Solution :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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