Straight Line Question 247
Through the point $ P(\alpha ,\beta ) $ , where $ \alpha \beta >0. $ the straight line $ \frac{x}{a}+\frac{y}{b}=1 $ is drawn so as to form with the axes a triangle of area S. If $ ab>0, $ then least value of S is
Options:
A) $ \alpha \beta $
B) $ 2\alpha \beta $
C) $ 3\alpha \beta $
D) None of these
Correct Answer: B $ \Rightarrow \frac{\alpha }{a}+\frac{\alpha \beta }{2S}=1 $ [using (i)] $ \therefore a\in \mathbb{R}\Rightarrow D\ge 0 $ $ 4S^{2}-8\alpha \beta S\ge 0 $ $ \Rightarrow S\ge 2\alpha \beta . $ Least value of $ S=2\alpha \beta . $
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Answer:
Solution:
$ \Rightarrow a^{2}\beta -2aS+2aS=0 $
BETA
