Question

If the bisectors of the angles between the pairs of lines given by the equation $a{x}^2+2hxy+b{y}^2=0$ and $a{x}^2+2hxy+b{y}^2+\lambda \left(x^2+y^2\right)=0$ be coincident, then $\lambda =$

• A. $a$
• B. $b$
• C. $h$
• D. Any real number

Answer 1

The correct option is D

Any real number

Explanation for the correct option.

Given pair of lines, $a{x}^2+2hxy+b{y}^2=0....\left(1\right)$ and $a{x}^2+2hxy+b{y}^2+\lambda \left(x^2+y^2\right)=0...\left(2\right)$.

The angle bisector of (1) is given by $\frac{\left(x^2-y^2\right)}{(a-b)}=\frac{xy}{h}$
$\Rightarrow h{x}^2-h{y}^2-(a-b)xy=0....\left(3\right)$
The angle bisector of equation (2) is given by:

$\frac{\left(x^2-y^2\right)}{(A-B)}=\frac{xy}{H}$; where $A=a+\lambda ,H=h,B=b+\lambda$

$\begin{array}{rcl}h{x}^2-h{y}^2-(a+\lambda -b-\lambda )xy& =& 0\\ h{x}^2-h{y}^2-(a-b)xy& =& 0.....\left(4\right)\end{array}$
By equations (3) and (4) are the same for any value of $\lambda$.
Hence, option D is correct.