Question

$P:2x^2-axy+6y^2=0$
$Q:3x^2-8xy+4y^2=0$

The angle between the lines given by $P$ is same as the angle between the lines given by $Q$ for

• A. $a=1$
• B. $a=-1$
• C. $a=\frac{8}{7}$
• D. no real value of $a$

Answer 1

The correct option is D no real value of $a$

For $2x^2-axy+6y^2=0$
$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{{(-\frac{a}{2})}^2-2.6}}{2+6}=\frac{\sqrt{{(-\frac{a}{2})}^2-2.6}}{4}$ ...(1)
And for $3x^2-8xy+4y^2=0$
$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{{(-4)}^2-3.4}}{3+7}=\frac{2}{5}$ ...(2)
from (1) and (2)
$\frac{\sqrt{{(-\frac{a}{2})}^2-2.6}}{4}=\frac{2}{5}\Rightarrow \sqrt{{(-\frac{a}{2})}^2-2.6}=\frac{8}{5}\Rightarrow {(-\frac{a}{2})}^2-18=\frac{64}{25}\Rightarrow \frac{a^2}{4}=-\frac{193}{2}$