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 $aʹx^2+2hʹxy+bʹy^2+2gʹx=0$
will be at right angles, if

• A. $(aʹ+bʹ)gʹ=(a+b)g$
• B. $(a+b)gʹ=(aʹ+bʹ)g$
• C. $h^2-ab=h^{\prime 2}-a^{\prime }b$
• D. $a+b+h^2=aʹ+bʹ+h{ʹ}^2$

Answer 1

Answer: (B)

The pair of lines joining origin to the point of intersection of two curves is a homogeneous equation.
The intersection of two curves
$a{x}^2+2hxy+b{y}^2+2gx+\lambda (a^{{}^{\prime }}{x}^2+2h^{{}^{\prime }}xy<br/><br/>+b^{{}^{\prime }}{y}^2+2g^{{}^{\prime }}x)=0$
$\Rightarrow x^2\left(a+a^{{}^{\prime }}\lambda \right)+2xy\left(h+h^{{}^{\prime }}\lambda \right)+y^2\left(b+\lambda b^{{}^{\prime }}\right)<br/><br/>+2x\left(g+\lambda g^{{}^{\prime }}\right)=0$
For making homogeneous equation, $g+\lambda g^{{}^{\prime }}=0$
$\Rightarrow \lambda =-\frac{g}{g^{{}^{\prime }}}$
Since, lines are perpendicular.
$\therefore$ Coefficient of $x^2+$ Coefficient of $y^2=0$
$\Rightarrow a+a^{{}^{\prime }}\lambda +b+b^{{}^{\prime }}\lambda =0$
$\Rightarrow a+b=-\left(a^{{}^{\prime }}+b^{{}^{\prime }}\right)\left(-\frac{g}{g^{{}^{\prime }}}\right)$
$\Rightarrow (a+b)g^{{}^{\prime }}=\left(a^{{}^{\prime }}+b^{{}^{\prime }}\right)g$