Q114 / 180 · COMEDK 2022 · Mathematics
Let the equation of the pair of lines $y=p x$ and $y=q x$ can be written as $(y-p x)(y-q x)=0$. Then the equation of the pair of the angle bisectors of the line $x^2-4 x y-5 y^2=0$ is
- A$x^2-3 x y+y^2=0$
- B$x^2+4 x y-y^2=0$
- Correct answerC$x^2+3 x y-y^2=0$
- D$x^2-3 x y-y^2=0$
Solution
The given equation of the pair of lines is $x^2 - 4xy - 5y^2 = 0$.
Comparing this with the general homogeneous equation of the second degree $ax^2 + 2hxy + by^2 = 0$, we have $a = 1$, $2h = -4$ (so $h = -2$), and $b = -5$.
The equation of the pair of angle bisectors is given by the formula $\dfrac{x^2 - y^2}{a - b} = \dfrac{xy}{h}$.
Substituting the values of $a$, $b$, and $h$ into the formula:
$\dfrac{x^2 - y^2}{1 - (-5)} = \dfrac{xy}{-2}$
$\dfrac{x^2 - y^2}{6} = \dfrac{xy}{-2}$
Multiplying both sides by 6:
$x^2 - y^2 = -3xy$
$x^2 + 3xy - y^2 = 0$.
Answer: $x^2+3 x y-y^2=0$
Comparing this with the general homogeneous equation of the second degree $ax^2 + 2hxy + by^2 = 0$, we have $a = 1$, $2h = -4$ (so $h = -2$), and $b = -5$.
The equation of the pair of angle bisectors is given by the formula $\dfrac{x^2 - y^2}{a - b} = \dfrac{xy}{h}$.
Substituting the values of $a$, $b$, and $h$ into the formula:
$\dfrac{x^2 - y^2}{1 - (-5)} = \dfrac{xy}{-2}$
$\dfrac{x^2 - y^2}{6} = \dfrac{xy}{-2}$
Multiplying both sides by 6:
$x^2 - y^2 = -3xy$
$x^2 + 3xy - y^2 = 0$.
Answer: $x^2+3 x y-y^2=0$
Hence, the correct answer is option: 'C'.