Find the product of the perpendiculars drawn from the point $\left(x_1,y_1\right)$ on the lines $a{x}^2+2hxy+b{y}^2=0$.

 

 Let the lines represented by $a{x}^{2}2hxyb{y}^2=0$ be $y=m_{1}x$ and $y=m_{2}x$.

Therefore, $m_1{m}_2=-\frac{2h}{b}$ and $m_1{m}_2=\frac{a}{b}$

The lengths of the perpendiculars from $\left(x_1,y_1\right)$ on these lines are

$\frac{\left|y_1-m_1{x}_1\right|}{\sqrt{1+m_1^2}}\text{ and }\frac{\left|y_1-m_2{x}_1\right|}{\sqrt{1+m_2^2}}$

Their product $=\frac{\left|y_1-m_1{x}_1\right|}{\sqrt{1m_1^2}}\times \frac{\left|y_1-m_2{x}_1\right|}{\sqrt{1m_2^2}}$

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

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

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

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

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

hence, the correct answer is option 1.