Question

The area of the triangle formed by the pair of straight lines $(a x+b y)^{2}-3(b x-a y)^{2}=0$ and $a x+b y+c=0$ is

• A. $\frac{c^{2}}{a^{2}+b^{2}}$
• B. $\frac{c^{2}}{2\left(a^{2}+b^{2}\right)}$
• C. $\frac{c^{2}}{\sqrt{2}\left(a^{2}+b^{2}\right)}$
• D. $\frac{c^{2}}{\sqrt{3}\left(a^{2}+b^{2}\right)}$

Answer 1

Answer: (D)

Given that, equation of pair of straight lines is
$(a x+b y)^{2}-3(b x-a y)^{2}=0$
$\Rightarrow a^{2} x^{2}+b^{2} y^{2}+2 a b x y$
$-3\left(b^{2} x^{2}+a^{2} y^{2}-2 a b x y\right)=0$
$\Rightarrow \left(a^{2}-3 b^{2}\right) x^{2}-\left(b^{2}-3 a^{2}\right) y^{2}+8 a b x y=0$
Let the equation of line be $y=m_{1} \,x,\, y=m_{2} \,x$.
$\therefore m_{1}+m_{2}=\frac{-4 a b}{b^{2}-3 a^{2}}$
$\begin{bmatrix}m_{1}+m_{2}=-\frac{2h}{b}\\ \because\\ m_{1} m_{2}=\frac{a}{b}\end{bmatrix}$
and $m_{1} m_{2}=\frac{a^{2}-3 b^{2}}{b^{2}-3 a^{2}}$
Now, $\left(m_{1}-m_{2}\right)^{2}=\left(m_{1}+m_{2}\right)^{2}-4 m_{1}\, m_{2}$
$=\frac{16 a^{2} b^{2}}{\left(b^{2}-3 a^{2}\right)^{2}}-\frac{4\left(a^{2}-3 b^{2}\right)}{\left(b^{2}-3 a^{2}\right)}$
$=\frac{4}{\left(b^{2}-3 a^{2}\right)^{2}} \left[4 a^{2} b^{2}-\left(a^{2}-3 b^{2}\right)\left(b^{2}-3 a^{2}\right)\right]$
$=\frac{4}{\left(b^{2}-3 a^{2}\right)^{2}}\left[4 a^{2} b^{2}-\left(a^{2} b^{2}-3 a^{4}\right.\right.\left.\left.-3 b^{4}+9 a^{2} b^{2}\right)\right]$
$=\frac{4}{\left(b^{2}-3 a^{2}\right)^{2}}\left[-6 a^{2} b^{2}+3 a^{4}+3 b^{4}\right]$
$=\frac{24}{\left(b^{2}-3 a^{2}\right)^{2}}\left(a^{2}-b^{2}\right)^{2}$
$\therefore \left(m_{1}-m_{2}\right)=\frac{2 \sqrt{3}}{\left(b^{2}-3 a^{2}\right)} \cdot\left(a^{2}-b^{2}\right)\,\,\,...(i)$
The intersecting point on the lines $y=m_{1}\, x$,
$y=m_{2} x$ and $a x+b y+c=0$
are $A(0,0), B\left(\frac{-c}{a+b m_{1}}, \frac{-c m_{2}}{a+b m_{1}}\right)$
and $C\left(\frac{-c}{a+b m_{2}}, \frac{-c m_{1}}{a+b m_{2}}\right)$
Area of $\Delta=\frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ \frac{-c}{a+b m_{1}} & \frac{-c m_{1}}{a+b m_{1}} & 1 \\ \frac{-c}{a+b m_{2}} & \frac{-c m_{2}}{a+b m_{2}} & 1\end{vmatrix}$
$=\frac{1}{2}\left[\frac{+c^{2} m_{2}}{\left(a+b m_{1}\right)\left(a+b m_{2}\right)}-\frac{c^{2} m_{1}}{\left(a+b m_{1}\right)\left(a+b m_{2}\right)}\right]$
$=-\frac{1}{2}\left[\frac{c^{2}\left(m_{1}-m_{2}\right)}{a^{2}+a b\left(m_{1}+m_{2}\right)+b^{2} m_{1} m_{2}}\right]$
$=\frac{-\frac{1}{2}\left[\frac{c^{2} 2 \sqrt{3}}{\left(b^{2}-3 a^{2}\right)} \cdot\left(a^{2}-b^{2}\right)\right]}{a^{2}+a b\left(\frac{-4 a b}{b^{2}-3 a^{2}}\right)+b^{2}\left(\frac{a^{2}-3 b^{2}}{b^{2}-3 a^{2}}\right)}[$ from Eq.(i)]
$=\frac{-\sqrt{3}\left[\left(a^{2}-b^{2}\right) c^{2}\right]}{a^{2}\left(b^{2}-3 a^{2}\right)-4 a^{2} b^{2}+a^{2} b^{2}-3 b^{4}}$
$=\frac{-\sqrt{3}\left[c^{2}\left(a^{2}-b^{2}\right)\right]}{a^{2} b^{2}-3 a^{4}-3 a^{2} b^{2}-3 b^{4}}$
$=\frac{-\sqrt{3} c^{2}}{-3\left(a^{4}+b^{4}+2 a^{2} b^{2}\right)}$
$=\frac{c^{2}}{\sqrt{3}\left(a^{2}+b^{2}\right)^{2}}$