Question

The condition that one of the lines represented by $a{x}^2+2hxy+b{y}^2=0$ is perpendicular to one of the lines represented by $a_1{x}^2+2h_{1}xy+b_1{y}^2=0$ is

Answer 1

$\begin{array}{c}Accordingtotheequation:\\ a{x}^2+2hxy+b{y}^2=0,a_1{x}^2+2h_{1}xy+b_1{y}^2=0\\ lets,slopeofequation:\\ m,m_1,m,m_2\\ sumofroots:\\ m+m_1=\frac{-2h}{b},m+m_2=\frac{-2h_1}{b_1}\\ \Rightarrow m.m_1=\frac{a}{b},m_1=\frac{a}{bm},\Rightarrow m.m_2=\frac{a_1}{b_1},m_2=\frac{a_1}{b_{1}m}\\ \Rightarrow m+\frac{a}{bm}=\frac{-2h}{b},\Rightarrow m+\frac{a_1}{b_{1}m}=\frac{-2h_1}{b_1}\\ m-\frac{2h}{b}+\frac{a}{bm}=0---\left(i\right),m-\frac{2h^{}}{b_1}+\frac{a_1}{b_{1}m}=0---\left(ii\right)\\ Now,wholeequmultiplybybm.multiplyby{b}_{1}m\\ b{m}^2+2hm+a=0,b_1{m}^2-2h_{1}m+a_1=0\\ constantlineslop:\\ m=\frac{-2h+\sqrt{4h^2-4ab}}{2b},m=\frac{-2h_1+\sqrt{4{h_1}^2-4a_1{b}_1}}{2b_1}\\ m=\frac{-h+\sqrt{h^2-ab}}{b},m=\frac{-h_1+\sqrt{{h_1}^2-a_1{b}_1}}{b_1}\\ bothequarerepresentoneline:\\ \Rightarrow \frac{-h+\sqrt{h^2-ab}}{b}=\frac{-h_1+\sqrt{{h_1}^2-a_1{b}_1}}{b_1}\\ \Rightarrow -h{b}_1+b_1\sqrt{h^2-ab}=-h_{1}b+b\sqrt{{h_1}^2-a_1{b}_1}\\ Now,squaringbothside,\\ {\left(h_{1}b-h{b}_1\right)}^2={\left(b\sqrt{{h_1}^2-a_1{b}_1}-b_1\sqrt{h^2-ab}\right)}^2\\ \Rightarrow {h_1}^2{b}^2+h^2{b_1}^2-2h_{1}hb{b}_1=b^2\left({h_1}^2-a_1{b}_1\right)+{b_1}^2\left(h^2-ab\right)-2b{b}_1\left(\sqrt{{h_1}^2-a_1{b}_1}\sqrt{h^2-ab}\right)\\ \Rightarrow {h_1}^2{b}^2+h^2{b_1}^2-2h_{1}hb{b}_1=b^2{h_1}^2-b^2{a}_1{b}_1+{b_1}^2{h}^2-{b_1}^{2}ab-2b{b}_1\left(\sqrt{{h_1}^2-a_1{b}_1}\sqrt{h^2-ab}\right)\\ \Rightarrow -2h_{1}hb{b}_1=-b^2{a}_1{b}_1-{b_1}^{2}ab-2b{b}_1\left(\sqrt{({h_1}^2-a_1{b}_1)(h^2-ab)}\right)\\ \Rightarrow -2h_{1}hb{b}_1=-b^2{a}_1{b}_1-{b_1}^{2}ab-2b{b}_1\left(\sqrt{({h_1}^2{h}^2-{h_1}^{2}ab-h^2{a}_1{b}_1+a_1{b}_{1}ab)}\right)\\ \Rightarrow b^2{a}_1{b}_1+{b_1}^{2}ab-2h_{1}hb{b}_1=-2b{b}_1\left(\sqrt{({h_1}^2{h}^2-{h_1}^{2}ab-h^2{a}_1{b}_1+a_1{b}_{1}ab)}\right)\\ \Rightarrow \left((a_{1}b+a{b}_1)-\left(2h_{1}h\right)\right)=-2\left(\sqrt{({h_1}^2{h}^2-{h_1}^{2}ab-h^2{a}_1{b}_1+a_1{b}_{1}ab)}\right)\\ wholesquaringbothside:\\ \Rightarrow {\left[(a_{1}b+a{b}_1)-\left(2h_{1}h\right)\right]}^2={\left[-2\left(\sqrt{({h_1}^2{h}^2-{h_1}^{2}ab-h^2{a}_1{b}_1+a_1{b}_{1}ab)}\right)\right]}^2\\ \Rightarrow {a_1}^2{b}^2+a^2{b_1}^2+4{h_1}^2{h}^2+2a{a}_{1}b{b}_1-4a_{1}b{h}_{1}h-4a{b}_1{h}_{1}h=4\left({h_1}^2{h}^2-{h_1}^{2}ab-h^2{a}_1{b}_1+a_1{b}_{1}ab\right)\\ \Rightarrow {a_1}^2{b}^2+a^2{b_1}^2+4{h_1}^2{h}^2+2a{a}_{1}b{b}_1-4a_{1}b{h}_{1}h-4a{b}_1{h}_{1}h=4{h_1}^2{h}^2-4{h_1}^{2}ab-4h^2{a}_1{b}_1+4a_1{b}_{1}ab\\ \Rightarrow {a_1}^2{b}^2+a^2{b_1}^2+2a{a}_{1}b{b}_1-4a_{1}b{h}_{1}h-4a{b}_1{h}_{1}h=-4{h_1}^{2}ab-4h^2{a}_1{b}_1+4a_1{b}_{1}ab\\ \therefore {\left(a_{1}b+a{b}_1\right)}^2=4\left(a{h}_1-a_{1}h\right)\left(h{b}_1-h^{1}b\right).prove\\ Sothatoneofthelineiscoincident.\end{array}$