EAMCET Engineering 2005 Solved Paper

Given that$12 x^2+25 x y+12 y^2+10 x+11 y+2=0$ . . . (i)First we take homogeneous part of Eq. (i),i.e. $12 x^2+25 x y+12 y^2=0$$\Rightarrow (3 x+4 y)(4 x+3 y)=0$So, let the lines represented by Eq. (i) be$\;3 x+4 y+c_1=0$ . . . (ii) $\;4 x+3 y+c_2=0$ . . . (iii)The combined Eqs. of (ii) and (iii), we get$\; (3 x+4 y+c_1)(4 x+3 y+c_2)=0$$\Rightarrow (3 x+4 y)(4 x+3 y)+\;c_1(4 x+3 y)+$$\;c_2(3 x+4 y)+c_1 c_2=0$$\Rightarrow \;\; 12 x^2+25 x y+12 y^2+\; (4 c_1+3 c_2) x$$\;+ (3 c_1+4 c_2) y+c_1 c_2=0$On comparing the equation with Eq. (i), we get$\;4 c_1+3 c_2=10$ . . . (iv)$\;3 c_1+4 c_2=11$ . . . (v)On solving equations (iv) and (v), we get $\;4 c_1+3 c_2=10$$\;3 c_1+4 c_2=11$On solving equations (iv) and (v), we get$c_1=1 \;\text{ }\; c_2=2$Separate equation of lines are$\;\; \;3 x+4 y+1=0$ . . . (vi) and $\;4 x+3 y+2=0$ . . . (vii)The perpendicular distance from origin to the equations (vi) and (vii) are$\;p_1=\;\frac{|0+0+1|}{\sqrt{3^2+4^2}}=\;\frac{1}{5}\;\text{ and }\; p_2=\;\frac{|0+0+2|}{\sqrt{4^2+3^2}}=\;\frac{2}{5}$$\therefore \;p_1 \cdot p_2=\;\frac{1}{5} \cdot \;\frac{2}{5}=\;\frac{2}{25}$