Line `x/a+y/b=1` cuts the co-ordinate axes at A(a,0) and B(0,b) and the line `x/a'+y/b'=-1` at `A'(-a',0)` and `B'(0,-b')`. If the points A,B,A',B' are concyclic then the orthocentre of triangle ABA' is

Options

A. `(0, 0)`
B. `(0, b')`
C. `(0, (aa')/(b))`
D. `(0, (b b')/(a))`

Answer

B, C

Explanation

If `A, B, A', B'` are concyclic then
`rArr (OA)(OA') = (OB)(OB')`
`rArr -b b' = -aa'`
For `DeltaABA'` orthocenter lies on the `OB`
`:.` Let the co-odinate of orthocentre is `H(0,k)` then
Slope of `A'H rArr ((k)/(a'))`
Slope of `AB rArr (-(b)/(a))`
`:. ((k)/(a'))(-(b)/(a)) = -1`
`:.` co-ordinate of orthocenter are `(0, b')(0, (a a')/(b))`