Question

The product of the perepndiculars drawn from the point $(x_1,y_1)$ on the lines $a{x}^2+2hxy+b{y}^2=0$ is

• A. $\frac{a{x_1}^2+2h{x}_1{y}_1+b{y_1}^2}{\sqrt{{(a-b)}^2+4h^2}}$
• B. $\frac{|a{x_1}^2+2h{x}_1{y}_1+b{y_1}^2|}{\sqrt{{(a-b)}^2+4h^2}}$
• C. $\frac{a{x_1}^2-2h{x}_1{y}_1+b{y_1}^2}{\sqrt{{(a-b)}^2+4h^2}}$
• D. $\frac{|a{x_1}^2-2h{x}_1{y}_1+b{y_1}^2|}{\sqrt{{(a-b)}^2+4h^2}}$

Answer 1

The correct option is B $\frac{|a{x_1}^2+2h{x}_1{y}_1+b{y_1}^2|}{\sqrt{{(a-b)}^2+4h^2}}$

Let $y=m_{1}x$ and $y=m_{2}x$ be the two lines given by $x^2+2hxy+b{y}^2=0$ so that

$m_1+m_2=\frac{-2h}{b}$ and $m_1{m}_2=\frac{a}{b}$ ...(1)

The product of the perpendiculars drawn from $\left(x_{1,}{y}_1\right)$ on these lines

$=\frac{|y_1-m_1{x}_1|}{\sqrt{1+{m_1}^2}}.\frac{|y_1-m_2{x}_1|}{\sqrt{1+{m_2}^2}}$

$=\frac{{y_1}^2-(m_1+m_2)x_1{y}_1+m_1{m}_2{x^2}_1}{\sqrt{1+{m^2}_1+{m_2}^2+{m_1}^2{m_2}^2}}$

$=$$\frac{\left|{y_1}^2-(m_1+m_2)x_1{y}_1+m_1{m}_2{x_1}^2\right|}{\sqrt{1+{(m_1+m_2)}^2-2m_1{m}_2+{m_1}^2{m_2}^2}}$

$=\frac{|{y_1}^2+\frac{2h{x}_1{y}_1}{b}+\frac{a{x_1}^2}{b}|}{\sqrt{1+\frac{4h^2}{b^2}-\frac{2a}{b}+\frac{a^2}{b^2}}}$ (using(1) )

$=\frac{|{a{x}_1}^2+2h{x}_1{y}_1+{{by}_1}^2|}{\sqrt{{(a-b)}^2+4h^2}}$