Question

Condition for which the slope of a line represented by $a{x}^2+2hxy+b{y}^2=0$ is square of the other is ,

• A. $ab(a+b-6h)+8h^3=0$
• B. $ab(a+b-6h)=8h^3$
• C. $ab(a+b+6h)+8h^3=0$
• D. None of thses

Answer 1

The correct option is A $ab(a+b-6h)+8h^3=0$

According to question.......

$\begin{array}{c}a{x}^2+2hxy+b{y}^2=0\\ ifwehavetwoliney=m_{1}x\&y=m_{2}xandthisisthesquareoftheother,m_2={m_1}^2\\ then,\\ y=m_{1}x\&y={m_1}^{2}x(wehavetwostraightlineeq{u}^n\\ know,\\ (y-m_{1}x)(y-{m_1}^{2}x)=0\\ \Rightarrow y^2-x(m_1+{m_1}^2)+{m_1}^3{x}^2=0\\ Now,compare:\\ m_1+{m_1}^2=\frac{-2h}{b},\&{m_1}^3=\frac{a}{b}\\ \Rightarrow (m_1+{m_1}^2{)}^3=\frac{-8h^3}{b^3}\\ \Rightarrow {m_1}^3+{m_1}^6+3m_1^3(m_1+{m_1}^2)=\frac{-8h^3}{b^3}\\ \Rightarrow \frac{a}{b}+\frac{a^2}{b^2}+3\left(\frac{a}{b}\right).\left(\frac{-2h}{b}\right)=\frac{-8h^3}{b^3}\\ \Rightarrow \frac{a}{b}+\frac{a^2}{b^2}+\frac{8h^3}{b^3}=\frac{6ah}{b^2}\\ \Rightarrow \frac{a}{b}\left[1+\frac{a}{b}\right]+\frac{8h^3}{b^3}=\frac{6ah}{b^2}\\ \Rightarrow \frac{a(a+b)}{b^2}+\frac{8h^3}{b^3}=\frac{6ah}{b^2}\\ \Rightarrow \frac{a(a+b)}{ah}+\frac{8h^3}{bah}=6\\ \Rightarrow \frac{a+b}{h}+\frac{8h^2}{ab}=6\\ \Rightarrow \frac{ab(a+b)+8h^3}{h\left(ab\right)}=6\\ \Rightarrow ab(a+b)+8h^3=6h\left(ab\right)\\ \therefore ab(a+b-6h)-8h^3=0\\ So,thatthecorrectoptionisA.\\ \end{array}$