Question

To the lines $a{x}^2+2hxy+b{y}^2=0$ , the line $a^2{x}^2+2h(a+b)xy+b^2{y}^2=0$ are

• A. equally inclined
• B. perpendicular
• C. bisector of the angle
• D. None of the above

Answer 1

Answer: (A)

The equation of the bisectors of the angle between the lines given by
$a{x}^2+2hxy+b{y}^2=0$ is
$\frac{x^2-y^2}{a-b}=\frac{xy}{h}...\left(i\right)$
And the equation of the bisectors of the angle between the lines given by
$a^2{x}^2+2h\left(a+b\right)xy+b^2{y}^2=0$ is
$\frac{x^2-y^2}{a^2-b^2}=\frac{xy}{h\left(a+b\right)}$
$\Rightarrow \frac{x^2-y^2}{a-b}=\frac{x}{h}...\left(ii\right)$
From eqs. $\left(i\right)$ and $\left(ii\right)$, it is clear that both the pair of straight lines have the same bisector, hence, the given two pairs of straight lines are equally inclined.