Question

If the normal form of the equation of a straight line $4x+3y+2=0$ is $x\cos \alpha +y\sin \alpha =p$ and its intercept form is $\frac{x}{a}+\frac{y}{b}=1$, then $\frac{p\sec \alpha }{ab}=$

• A. $-\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $-\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

Given, $4x+3y+2=0$
$\Rightarrow -4x-3y=2$
$\frac{-4x}{5}-\frac{3}{5}y=\frac{2}{5}$
$\because \cos \alpha =\frac{-4}{5},\sin \alpha =\frac{-3}{5}$ and $P=\frac{2}{5}$
$\frac{x}{2/-4}+\frac{y}{2/-3}=1$
$\Rightarrow a=\frac{-2}{4},b=\frac{-2}{3}$
$\therefore \frac{P\sec \alpha }{ab}=\frac{\frac{2}{5}\times \left(\frac{-5}{4}\right)}{\left(\frac{-2}{4}\right)\left(\frac{-2}{3}\right)}$
$=\frac{-3}{2}$