Question

The area of the triangle formed by the axes and the line $(\cosh \alpha -sinh\alpha )x+(\cosh \alpha +sinh\alpha )y=2$ in square units is :

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

Given, $(\cos h\alpha -\sin h\alpha )x+(\cos h\alpha +\sin h\alpha )y=2$

$e^{\alpha }x-e^{-\alpha }y=2$

$\because \cos h\alpha =\frac{e^{\alpha }+e^{-\alpha }}{2}$ & $\sinh \alpha =\frac{e^{\alpha }-e^{-\alpha }}{2}$

At $y=0,x=\frac{2}{e^{\alpha }}$

At $x=0,y=-2e^{\alpha }$

So, area of $\Delta ABC=\frac{1}{2}\times AC\times BC$

$=\frac{1}{2}\times 2e^{\alpha }\times \frac{2}{e^{\alpha }}$

$=2$