Question

If the bisectors of the angles between the pairs of lines given by the equation $ax^{2} + 2hxy + by^{2} = 0$ and

$ax^{2} + 2hxy + by^{2} + \lambda(x^{2} + y^{2}) = 0$ be coincident, then $\lambda =$

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

Answer 1

Answer: (D)

Any real number

Answer 2

Bisectors of $ax^{2} + 2hxy + by^{2} = 0$ are

$\frac{x^{2} - y^{2}}{a - b} = \frac{xy}{h}$ .....(i)

and of $ax^{2} + 2hxy + by^{2} + \lambda(x^{2} + y^{2}) = 0$

i.e., $(a + \lambda)x^{2} + 2hxy + (b + \lambda)y^{2} = 0$are

$\frac{x^{2} - y^{2}}{(a + \lambda) - (b + \lambda)} = \frac{xy}{h}$ .....(ii)

Which is the same equation as equation (i). Hence for any $\lambda$belonging to real numbers, the lines will have same bisectors.