SATHEE: Area Question 54

Area Question 54

Question 54

For any real $t, x = \frac{e^{t} + e^{- t}}{2}, y = \frac{e^{t} - e^{- t}}{2}$ is a point on the hyperbola $x^{2} - y^{2} = 1$ . Find the area bounded by this hrperbola and the lines joining its centre to the points corresponding to $t_{1}$ and $- t_{1}$ .

$(1982, 3 M)$

Show Answer

Solution:

Let $P = \frac{e^{t_{1}} + e^{- t_{1}}}{2}, \frac{e^{t_{1}} - e^{- t_{1}}}{2}$

and $Q = \frac{e^{- t} + e^{t_{1}}}{2}, \frac{e^{- t_{1}} - e^{t}}{2}$

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 $(- t_{1})$ .

Required area

$\begin{aligned} = 2 area of △ P C N - \int_{1}^{\frac{e^{t_{1}} + e^{- t_{1}}}{2}} y d x & = 2 \frac{1}{2} \frac{e^{t_{1}} + e^{- t_{1}}}{2} \frac{e^{t_{1}} - e^{- t_{1}}}{2} - \int_{1}^{t_{1}} y \frac{d y}{d t} \cdot d t & = 2 \frac{e^{2 t_{1}} - e^{- 2 t_{1}}}{8} - \int_{0}^{t_{1}} \frac{e^{t} - e^{- t}}{2} d t & = \frac{e^{2 t_{1}} - e^{- 2 t_{1}}}{4} - \frac{1}{2} \int_{0}^{t_{1}} (e^{2 t} + e^{- 2 t} - 2) d t & = \frac{e^{2 t_{1}} - e^{- 2 t_{1}}}{4} - \frac{1}{2} \frac{e^{2 t}}{2} - \frac{e^{- 2 t}}{2} - 2 t & = \frac{e^{2 t_{1}} - e^{- 2 t_{1}}}{4} - \frac{1}{4} (e^{2 t_{1}} - e^{- 2 t_{1}} - 4 t_{1}) \end{aligned}$

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