Question

If one of the lines given by the equation $2x^2+pxy+3y^2=0$ coincides with one of those given by $2x^2+qxy-3y^2=0$ and the other lines represented by them be perpendicular, then which of the following may be true ?

• A. $p=5$
• B. $p=-5$
• C. $q=-1$
• D. $q=1$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $p=5$
B $p=-5$
C $q=-1$
D $q=1$

Let
$\frac{2x^2}{3}+\frac{pxy}{3}+y^2=(y+mx)(y-ax)=0$ ...(i)
Therefore by comparing coefficients, from (i) we can get the following
$m-a=\frac{p}{3}$ ...(1)
$-ma=\frac{2}{3}$ ...(2)
Similarly,
$\frac{-2x^2}{3}+\frac{-qxy}{3}+y^2=(y+mx)(y+\frac{x}{a})=0$ ...(ii)
Therefore by comparing coefficients, from (ii) we get,
$m-\frac{1}{a}=\frac{-q}{3}$ ...(3)
$\frac{m}{a}=\frac{-2}{3}$ ...(4)
Hence from 3 we get
$\frac{am-1}{a}=\frac{-q}{3}$
Substituting the value of $m$ from equation 2 we get
$\frac{\frac{-2}{3}+1}{a}=\frac{-q}{3}$
$-aq=1$
$a=\frac{-1}{q}$
Hence, $m=\frac{2}{3q}$ and $m=\frac{2q}{3}$ from 4 and 2 respectively.
Therefore,
$\frac{2}{3q}=m=\frac{2q}{3}$
$q^2=1$
$q=\pm 1$
Hence $a=\pm 1$
Therefore $m=\pm \frac{2}{3}$
So $p=\pm 5$