Question

If $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents two straight lines equidistant from the origin, then $f^4-g^4$

• A. $b{f}^2-a{g}^2$
• B. $a{g}^2-b{f}^2$
• C. $c(b{f}^2-a{g}^2)$
• D. $c(a{g}^2-b{g}^2)$

Answer 1

The correct option is C $c(b{f}^2-a{g}^2)$

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$\text{let the twe lines be.}$

$L_1;y=m_{1}x+c_1$

$L_2:y=m_{2}x+c_2$

$(m_{1}x+c_1-y)(m_{2}x+c_2-y)=0-\text{(ii)}$

$\text{on comparing (i) and(ii) , we got}$

on comparing

(i) and (ii), we got

$m_1{c}_2+m_2{c}_1=\frac{2g}{b}$ & $c_1+c_2=-\frac{2f}{b}$

$m_2{m}_2$ $=\frac{a}{b}$ & $c_1{c}_2=\frac{c}{b}$

Now,

$\frac{\left|c_1\right|}{\sqrt{1+m_2^2}}=\frac{\mid c_{21}}{\sqrt{1+m_2^2}}$

Squaring both sides and cross-multiply

$\Rightarrow c_1^2+c_1^2{m}_2^2=c_2^2+c_2^2{m}_1^2$

$\Rightarrow (e_1^2-c_2^2)=(c_2^2{m}_2^2-c_1^2{m}_2^2)$

$\Rightarrow (c_1+c_2)(c_1-c_2)=(c_2{m}_1+c_1{m}_2)(c_2{m}_1-c_1{m}_2)$

$\Rightarrow (c_1+c_2)\sqrt{{(c_1+c_2)}^2-4c_1{c}_2}=(c_2{m}_1+c_1{m}_2)$

$\sqrt{{(E_2{m}_1+C_1{m}_2)}^2-4m_1{m}_2{C}_1{C}_2}$

$\Rightarrow -\frac{2f}{b}\sqrt{\frac{4f^2}{b^2}-\frac{4c}{b}}=\frac{2q}{b}\sqrt{\frac{4g^2}{b^2}-\frac{4a}{b}}$

$\Rightarrow g^2\left(\frac{g^2-ac}{b^2}\right)=f^2\left(\frac{f^2-bc}{b^2}\right)$

$\Rightarrow f^4-g^4=c(b{f}^2-a{g}^2)$