Through the point $$P\left( {\alpha ,\,\beta } \right),$$   where $$\alpha \beta > 0,$$   the straight line $$\frac{x}{a} + \frac{y}{b} = 1$$   is drawn so as to form with axes a triangle of area $$S.$$ If $$ab > 0,$$   then least value of $$S$$ is :

Solution :

Area of $$\Delta OAB = S = \frac{1}{2}ab......\left( {\text{i}} \right)$$
Equation of $$AB$$  is $$\frac{x}{a} + \frac{y}{b} = 1$$
Putting $$\left( {\alpha ,\,\beta } \right),$$  we get
$$\eqalign{ & \frac{\alpha }{a} + \frac{\beta }{b} = 1 \cr & \Rightarrow \frac{\alpha }{a} + \frac{{a\beta }}{{2S}} = 1\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{using }}\left( {\text{i}} \right)} \right] \cr & \Rightarrow {a^2}\beta - 2aS + 2\alpha S = 0 \cr & \therefore \,a\, \in \,R \Rightarrow D \geqslant 0 \cr & 4{S^2} - 8\alpha \beta S \geqslant 0 \cr & \Rightarrow S \geqslant 2\alpha \beta \cr} $$
Least value of $$S = 2\alpha \beta $$