Question

The two lines $l_1$x + $m_1$y + $n_1$ = $0$ and $l_2$x + $m_2$y + $n_2$ = $0$ are the conjugate lines w.r.t. to the hyperbola $S=0$ if:

• A. $a^3$ $l_1{l}_2$$+b^3$$m_1$$m_2$ = $n_1{n}_2$.
• B. $a^2$ $l_1{l}_2$$-b^2$$m_1$$m_2$ = $n_1{n}_2$.
• C. $a^2$ $l_1{l}_2$ $+b^2$$m_1$$m_2$ = -$n_1{n}_2$.
• D. None of these

Answer 1

The correct option is B $a^2$ $l_1{l}_2$$-b^2$$m_1$$m_2$ = $n_1{n}_2$.

Given: Two conjugate lines $l_{1}x+m_{1}y+n_1=0$----------1& $l_{2}x+m_{2}y+n_2=0$------------2

$S=0$ as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$----------3

From definition of Conjugate lines, Pole of line $L_1$ must lie on pole of $L_2$ and vice-versa.

Pole Equation 1: $(\frac{-a^2{l}_1}{n_1}+\frac{b^2{m}_1}{n_1})$

Pole Equation 2: $(\frac{-a^2{l}_2}{n_2}+\frac{b^2{m}_2}{n_2})$

So, $P_1$ must lie on Equation 2 and $P_2$ must lie on Equation 1

$l_1\cdot \left(\frac{-a^2{l}_2}{n_2}\right)+m_1\left(\frac{b^2{m}_2}{n_2}\right)+n_1=0$ and

$l_2\cdot \left(\frac{-a^2{l}_1}{n_1}\right)+m_2\left(\frac{b^2{m}_1}{n_1}\right)+n_2=0$

$\Rightarrow \frac{-a^2{l}_2{l}_1}{1}+\frac{b^2{m}_1{m}_2}{1}+n_1{n}_2=0$ and

$\Rightarrow \frac{-a^2{l}_1{l}_2}{1}+\frac{b^2{m}_1{m}_2}{1}+n_1{m}_2=0$

They both give the same result as

$\Rightarrow b^2{m}_1{m}_2+n_1{n}_2=a^2{l}_2{l}_1$

$\Rightarrow a^2{l}_2{l}_1-b^2{m}_1{m}_2=n_1{n}_2$.