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If the normal at four points P(i)(x(i), ...

If the normal at four points `P_(i)(x_(i), (y_(i)) l, I = 1, 2, 3, 4` on the rectangular hyperbola `xy = c^(2)` meet at the point `Q(h, k),` prove that
`x_(1) + x_(2) + x_(3) + x_(4) = h, y_(1) + y_(2) + y_(3) + y_(4) = k`
`x_(1)x_(2)x_(3)x_(4) =y_(1)y_(2)y_(3)y_(4) =-c^(4)`

Text Solution

Verified by Experts

The equation of the normal at `(ct, (c )/(t))` to the given hyperbola `xy = c^(2)` is given by
`t^(3)x -ty -ct^(4) + c = 0`
Which will through Q(h, k) if
`t^(3)h - -tk -ct^(4) + c =0`
`rArr ct^(4) - t^(3)h + tk -c =0`
which is a biquadratic equation in t giving four different value of t say `t_(1), t_(2), t_(3) and t_(4)` and hence corresponding four points `P_(i)(ct_(i), (c)/(t))` i.e., `P_(i)(x_(i), y_(i)), i - 1, 2, 3, 4`
We have `t_(1), + t_(2) + t_(3) + t_(4) = (h)/(c), t_(1)t_(2)t_(3) + t_(1)t_(3)t_(4) + t_(2)t_(3)t_(4) + t_(1)t_(2)t_(4) = (k)/(c)`
`rArr ct_(1) + ct_(2) + ct_(3) + ct_(4) = h and t_(1)t_(2)t_(3)t_(4) =-1`
` rArr x_(1) + x_(2) + x_(3) + x_(4) =h`
Also `(t_(1)t_(2)t_(3) + t_(1)t_(2)t_(4) + t_(1)t_(3)t_(4) + t_(2)t_(3)t_(4))/(t_(1)t_(2)t_(3)t_(4)) =(-(k)/(c))/(-1)`
`rArr (1)/(t_(1)) + (1)/(t_(2)) + (1)/(t_(3)) + (1)/(t_(4)) = (k)/(c)`
`rArr (c)/(t_(1)) + (c)/(t_(2)) + (c)/(t_(3)) + (c)/(t_(4)) =k`
`rArr y_(1) + y_(2)+ y_(3) + y_(4) = -1`
Morever, `t_(1)t_(2)t_(3)t_(4) =-1`
`rArr (ct_(1))(ct_(2))(ct_(3))(ct_(4)) =-c^(4) = x_(1)x_(2)x_(3)x_(4)`
and `(1)/(t_(1)t_(2)t_(3)t_(4)) =-1`
`rArr ((c)/(t_(1)))((c)/(t_(2)))((c)/(t_(3)))((c)/(t_(4)))=-c^(4)`
`rArr y_(1)y_(2)y_(3)y_(4) =-c^(4)`
Thus `x_(1)x_(2)x_(3)x_(4) = y_(1)y_(2)y_(3)y_(4) =-c^(4)`
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