Question

If one of the lines given by the equation $2{{x}^{2}}+axy+3{{y}^{2}}=0$ coincides with one of those given by $2{{x}^{2}}+bxy-3{{y}^{2}}=0$ and the other lines represented by them be perpendicular, then

(a) $a=-5,b=1$
(b) $a=5,b=-1$
(c) $a=5,b=1$
(d) None of these

Answer 1

Hint: Assume that the equation $2{{x}^{2}}+axy+3{{y}^{2}}=0$ represents the product of two lines of the form $y-mx$ and $y-m'x$. Use the fact that the product of the slope of two perpendicular lines is -1 to write the equation of lines whose product is
$2{{x}^{2}}+bxy-3{{y}^{2}}=0$. Simplify the equations to find the value of m and m’.
Substitute the values to calculate the values of ‘a’ and ‘b’.

Complete step-by-step answer:
We have two equations $2{{x}^{2}}+axy+3{{y}^{2}}=0$ and $2{{x}^{2}}+bxy-3{{y}^{2}}=0$, which represent the joint equation of lines, such that one line of both the curves coincide with each other and other lines represented by them are perpendicular to each other.

Dividing the equation $2{{x}^{2}}+axy+3{{y}^{2}}=0$ by 3, we have

$\dfrac{2{{x}^{2}}}{3}+\dfrac{axy}{3}+{{y}^{2}}=0$

Let’s assume that the equation $\dfrac{2{{x}^{2}}}{3}+\dfrac{axy}{3}+{{y}^{2}}=0$ represents

the lines $y-mx$ and $y-m'x$.

Thus, we have $\dfrac{2{{x}^{2}}}{3}+\dfrac{axy}{3}+{{y}^{2}}=\left( y-mx \right)\left( y-m'x

\right)$.

Simplifying the above equation, we have

$\dfrac{2{{x}^{2}}}{3}+\dfrac{axy}{3}+{{y}^{2}}={{y}^{2}}-mxy-m'xy+mm'{{x}^{2}}$.

Comparing the terms on both sides of the above equation, we have
$\dfrac{2}{3}=mm'......\left( 1 \right)$ and $-\left( m+m' \right)=\dfrac{a}{3}.....\left( 2

\right)$. We can rewrite equation (1) as $m'=\dfrac{2}{3m}.....\left( 3 \right)$.

Dividing the equation $2{{x}^{2}}+bxy-3{{y}^{2}}=0$ by -3, we have

$\dfrac{-2{{x}^{2}}}{3}-\dfrac{bxy}{3}+{{y}^{2}}=0$.

 Let’s assume that the line $y-mx$ is common to both the curves.

We know that the other line of the curve $\dfrac{-2{{x}^{2}}}{3}-\dfrac{bxy}{3}+{{y}^{2}}=0$

is perpendicular to the line $y-m'x$.

We know that the product of the slope of two perpendicular lines is -1.

Thus, the slope of the other line is $\dfrac{-1}{m'}$ . So, the equation of another line of

$\dfrac{-2{{x}^{2}}}{3}-\dfrac{bxy}{3}+{{y}^{2}}=0$ is $y+\dfrac{1}{m'}x$.

Thus, we have $\dfrac{-2{{x}^{2}}}{3}-\dfrac{bxy}{3}+{{y}^{2}}=\left( y-mx \right)\left(

y+\dfrac{1}{m'}x \right)$.

Simplifying the above equation, we have

$\dfrac{-2{{x}^{2}}}{3}-\dfrac{bxy}{3}+{{y}^{2}}={{y}^{2}}-mxy+\dfrac{xy}{m'}-\dfrac{m}{m'}{{x}

^{2}}$.

Comparing the terms on both sides, we have $\dfrac{-2}{3}=\dfrac{-m}{m'}.....\left( 4 \right)$

and $\dfrac{1}{m'}-m=\dfrac{-b}{3}.....\left( 5 \right)$.

Substituting equation (3) in equation (4), we have $\dfrac{-2}{3}=\dfrac{-m}{\dfrac{2}{3m}}$.

Simplifying the above equation, we have $\dfrac{4}{9}={{m}^{2}}$. Thus, we have $m=\pm

\dfrac{2}{3}$.

Substituting $m=\dfrac{2}{3}$ in equation (3), we have $m'=\dfrac{2}{3\left( \dfrac{2}{3}

\right)}=1$. Substituting these values in equation (2) and (5), we have $-\left( \dfrac{2}{3}+1

\right)=\dfrac{a}{3}\Rightarrow \dfrac{a}{3}=\dfrac{-5}{3}\Rightarrow a=-5$ and

$1-\dfrac{2}{3}=\dfrac{-b}{3}\Rightarrow \dfrac{1}{3}=\dfrac{-b}{3}\Rightarrow b=-1$.

Substituting $m=-\dfrac{2}{3}$ in equation (3), we have $m'=\dfrac{2}{3\left( \dfrac{-2}{3}

\right)}=-1$. Substituting these values in equation (2) and (5), we have $-\left( \dfrac{-2}{3}-1

\right)=\dfrac{a}{3}\Rightarrow \dfrac{a}{3}=\dfrac{5}{3}\Rightarrow a=5$ and

$-1+\dfrac{2}{3}=\dfrac{-b}{3}\Rightarrow \dfrac{-1}{3}=\dfrac{-b}{3}\Rightarrow b=1$.

Hence, the possible values of a and b are $a=5,b=1$, which is option (c).


Note: It’s necessary to consider all possible values of a and b. We can also solve this question by taking the equation of lines of the form $cx+dy$ and $c'x+d'y$. Write the equations and simplify them to find the value of ‘a’ and ‘b’.