Question

Perpendiculars $AL,AM$ are drawn from any point on the $x$-axis to the pair of lines $2x^2-5xy-3y^2=0$. The angle made by $LM$ with $+ve$ direction of $x$-axis is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$\frac{\pi }{4}$

$2x^2-5xy-3y^2=0$
$t=\frac{x}{y}$
$2t^2-5t-300$
$t=\frac{5\pm \sqrt{25+24}}{4}$
$t=3,\frac{-1}{2}$
$\Rightarrow y=\frac{x}{3}$ ----(1) & $y=-2x$ ----(2)are two seperate lines
Let, $A(h,0)$ is any point on $x-$axis.
$\therefore$ equation of lines passing through $(h,0)$ and perpendiculor to $y=\frac{x}{3}$ & $y=-2x$ are
$y=-3(x-h)$ & $y=\frac{1}{2}(x-h)$
$y+3x=3h$ ---(3) & $y-\frac{1}{2}x+\frac{h}{2}=0$ ---(4)
So, From 1 & 3, $x=\frac{9h}{10},y=\frac{3h}{10}\Rightarrow L(\frac{9h}{10},\frac{3h}{10})$
From 2 & 4, $x=\frac{h}{5},y=-2h5\Rightarrow M(\frac{h}{5},\frac{-2h}{5})$
$\therefore$ angle made by LM with +x axis $=\left(\frac{\frac{7h}{10}}{\frac{7h}{10}}\right)=1$
$tan\theta =1$
$\Rightarrow \theta =\frac{\pi }{4}$