Question

The pair of lines joining origin to the points of intersection of the two curves
$ ax^2 + 2hxy + by^2 + 2gx = 0 $
and $ aʹ \,x^2 + 2hʹ\, xy + bʹ\, y^2 + 2gʹ\, x = 0 $
will be at right angles, if

• A. $ (aʹ + bʹ )g ʹ = (a + b)g $
• B. $ (a + b)g ʹ = (aʹ + bʹ )g $
• C. $ h^2 - ab = h'^2 - a' \,b $
• D. $ a + b + h^2 = aʹ + bʹ + hʹ^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}+2 h x y+b y^{2}+2 g x +\lambda\left(a^{'} x^{2}+2 h^{'} x y\right.
+\left.b^{'} y^{2}+2 g^{'} x\right)=0 $
$\Rightarrow x^{2}\left(a+a^{'} \lambda\right)+2 x y\left(h+h^{'} \lambda\right)+y^{2}\left(b+\lambda b^{'}\right)
+ 2 x\left(g+\lambda g^{'}\right)=0 $
For making homogeneous equation, $g+\lambda g^{'}=0$
$\Rightarrow \lambda=-\frac{g}{g^{'}}$
Since, lines are perpendicular.
$\therefore $ Coefficient of $x^{2}+$ Coefficient of $y^{2}=0$
$\Rightarrow a+a^{'} \lambda+b+b^{'} \lambda=0$
$\Rightarrow a+b=-\left(a^{'}+b^{'}\right)\left(-\frac{g}{g^{'}}\right)$
$\Rightarrow (a+b) g^{'}=\left(a^{'}+b^{'}\right) g$