AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

The given equation of lines are, $3x^{2}+2hxy-3y^{2}=0$ And $3x^{2}+2hxy-3y^{2}+2x-4y+c=0$ The point of intersection of the lines given by $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0\;\text{ is }\;\left(\frac{hf-bg}{ab-h^{2}},\frac{gh-af}{ab-h^{2}}\right)$ Thus, the point of intersection of the second line is $A(x,y)=A\left(\frac{2h-3}{9+h^{2}},-\frac{h+6}{9+h^{2}}\right)$ The equation of diagonal of square passes through origin is, $y=\frac{h+6}{3-2h}x$ The equation of diagonal of square that does passes through the origin is $2x-4y+c=0$ The diagonal are perpendicular if, $\frac{1}{2}\left(\frac{h+6}{3-2h}\right)=-1$ $h+6\;\;=4h-6$ $3h\;\;=12$ $h\;\;=4$ Thus, $\text{Point}\;A\;\;=\left(\frac{8-3}{9+16},-\frac{10}{9+16}\right)$ $\;\;=\left(\frac{1}{5},-\frac{2}{5}\right)$ Consider the figure shown below. The midpoint of line OA is, $M=\left(\frac{1}{10},-\frac{1}{5}\right)$ Therefore, $\frac{2}{10}+\frac{4}{5}+c=0$ $c=-1$ $\;\text{Therefore,}\;(h,c)=(4,-1)$