Question

The line joining the origin to the points of intersection of the curves
$a{x}^2+2hxy+b{y}^2+2gx=0$ and $a^{{}^{\prime }}{x}^2+2h^{{}^{\prime }}xy+b^{{}^{\prime }}{y}^2+2g^{{}^{\prime }}x=0$
will be mutually perpendicular, if

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The family of lines passing through point of intersection of the given curves will be

$(a{x}^2+2hxy+b{y}^2+2gx=0)+$$\lambda (a^{{}^{\prime }}{x}^2+2h^{{}^{\prime }}xy+b^{{}^{\prime }}{y}^2+2g^{{}^{\prime }}x)=0$ $\Rightarrow (a+a^{{}^{\prime }}\lambda )x^2+(2h+2h^{{}^{\prime }}\lambda )xy+(b+b^{{}^{\prime }}\lambda )y^2+(2g+2g^{{}^{\prime }}\lambda )x=0$

Now the condition for perpendicularity is $△=0$ and a + b = 0.

$\Rightarrow a+a^{{}^{\prime }}\lambda +b+b^{{}^{\prime }}\lambda =0\Rightarrow \lambda =-\frac{a+b}{a^{{}^{\prime }}+b^{{}^{\prime }}}$

and $△=abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

$\Rightarrow 0+0-0-(b+b^{{}^{\prime }}\lambda )(2g+2g^{{}^{\prime }}\lambda {)}^2-0=0$

$\Rightarrow 4(b+b^{{}^{\prime }}\lambda )(g+g^{{}^{\prime }}\lambda {)}^2=0$

Now on putting the value of $\lambda$, we get

$g(a^{{}^{\prime }}+b^{{}^{\prime }})=g^{{}^{\prime }}(a+b)$.