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For any real t ,x=1/2(e^t+e^(-t)),y=1/2(...

For any real `t ,x=1/2(e^t+e^(-t)),y=1/2(e^t-e^(-t))` is a point on the hyperbola `x^2-y^2=1` Show that the area bounded by the hyperbola and the lines joining its centre to the points corresponding to `t_1a n d-t_1` is`t_1dot`

Text Solution

Verified by Experts

The correct Answer is:
`(e^(2t_(1))-e^(-2t_(1)))/(4)-(1)/(4) (e^(2t_(1))-e^(-2t_(1))-4t_(1))`
Let `P=((e^(t_(1))+e^(-t_(1)))/(2),(e^(t_(1))-e^(-t_(1)))/(2))`
`and Q=((e^(-t)+e^(t_(1)))/(2),(e^(-t_(1))-e^(t))/(2))`
We have to find the area of the region bounded by the curve `x^(2)-y^(2)=1` and the lines joining the centre `x=0, y=0 ` to the points `(t_(1)) and (-t_(1)).`

`=2["area of "Delta PCN -int_(1)^((e^(t_(1))+e^(-t_(1)))/(2))ydx]`
`=2[(1)/(2)((e^(t_(1))+e^(-t_(1)))/(2))((e^(t_(1))-e^(-t_(1)))/(2))-int_(1)^(t_(1))y (dy)/(dx)*dt]`
`=2[(e^(2t_(1))-e^(-2t_(1)))/(8)-int_(0)^(t_(1))((e^(t)-e^(-t))/(2))dt]`
`=(e^(2t_(1))-e^(-2t_(1)))/(4)-(1)/(2) int_(0)^(t_(1))(e^(2t)+e^(-2t)-2)dt`
`=(e^(2t_(1))-e^(-2t_(1)))/(4)-(1)/(2)[(e^(2t))/(2)-(e^(-2t))/(2)-2t]`
`=(e^(2t_(1))-e^(-2t_(1)))/(4)-(1)/(4) (e^(2t_(1))-e^(-2t_(1))-4t_(1))`
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