Question

The pair of lines joining origin to the points of intersection of the two curves $a{x}^2+2hxy+b{y}^2+2gx=0$ and $a\text{'}{x}^2+2h\text{'}xy+b\text{'}{y}^2+2g\text{'}x=0$ will be at right angles, if

• A. $\left(a\text{'}+b\text{'}\right)g\text{'}=\left(a+b\right)g$
• B. $\left(a+b\right)g\text{'}=\left(a\text{'}+b\text{'}\right)g$
• C. $h^2-ab=h{\text{'}}^2-a\text{'}b\text{'}$
• D. $a+b+h^2=a\text{'}+b\text{'}+h{\text{'}}^2$

Answer 1

The correct option is B

$\left(a+b\right)g\text{'}=\left(a\text{'}+b\text{'}\right)g$

Explanation for the correct option

Given equation of first curve is,

$$\begin{gathered} a{x}^2+2hxy+b{y}^2+2gx=0 \\ \Rightarrow 2x=-\frac{1}{g}\left(a{x}^2+2hxy+b{y}^2\right)...................\left(i\right) \end{gathered}$$

Given equation of second curve is,

$$\begin{gathered} a\text{'}{x}^2+2h\text{'}xy+b\text{'}{y}^2+2g\text{'}x=0 \\ \Rightarrow 2xg\text{'}=-\left(a\text{'}{x}^2+2h\text{'}xy+b\text{'}{y}^2\right)...................\left(ii\right) \end{gathered}$$

Putting $\left(i\right)$ in $\left(ii\right)$, we get,

$$\begin{gathered} \Rightarrow \left(a\text{'}{x}^2+2h\text{'}xy+b\text{'}{y}^2\right)=\left(\frac{g\text{'}}{g}\right)\left(a{x}^2+2hxy+b{y}^2\right) \\ \Rightarrow g\left(a\text{'}{x}^2+2h\text{'}xy+b\text{'}{y}^2\right)=g\text{'}\left(a{x}^2+2hxy+b{y}^2\right) \\ \Rightarrow x^2\left(ga\text{'}-ag\text{'}\right)+xy\left(2gh\text{'}-2hg\text{'}\right)+y^2\left(gb\text{'}-bg\text{'}\right)=0 \end{gathered}$$

Since it subtends right angle at origin, then the sum of the coefficient of $x^2$ and $y^2$ is $0$

$$\begin{gathered} \therefore \left(ga\text{'}-ag\text{'}\right)+\left(gb\text{'}-bg\text{'}\right)=0 \\ \Rightarrow g\left(a\text{'}+b\text{'}\right)-g\text{'}\left(a+b\right)=0 \\ \Rightarrow g\left(a\text{'}+b\text{'}\right)=g\text{'}\left(a+b\right) \end{gathered}$$

Hence, option $\left(\mathrm{B}\right)$ $\left(a+b\right)g\text{'}=\left(a\text{'}+b\text{'}\right)g$, is the correct answer.