Question

If the pair of lines $ax^2 - 7xy - 3y^2 = 0$ and $2x^2 + xy - 6y^2 = 0$ have exactly one line in common and '$a$' is an integer, then the equation of the pair of bisectors of the angles between the lines $ax^2 - 7xy - 3y^2 = 0$ is

• A. $7x^2 + 18xy - 7y^2 = 0$
• B. $x^2 - 16xy - y^2 = 0$
• C. $7x^2 - 9xy - 7y^2 = 0$
• D. $x^2 - 8xy - y^2 = 0$

Hint:

For pairs of lines with a common line, assume one line is shared and compare coefficients to find parameters. Use the angle bisector formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ for $ax^2 + 2hxy + by^2 = 0$.

Answer 1

The Correct Option is A

The second pair of lines is: \[ 2x^2 + xy - 6y^2 = (2x - 3y)(x + 2y) = 0 \] Lines are $2x - 3y = 0$ and $x + 2y = 0$. Suppose $2x - 3y = 0$ is common with $ax^2 - 7xy - 3y^2 = 0$. Then: \[ ax^2 - 7xy - 3y^2 = (2x - 3y)(\alpha x + \beta y) \] \[ = 2\alpha x^2 + (2\beta - 3\alpha)xy - 3\beta y^2 \] Compare coefficients with $ax^2 - 7xy - 3y^2$: \[ 2\alpha = a, \quad 2\beta - 3\alpha = -7, \quad -3\beta = -3 \] \[ \beta = 1, \quad 2 \cdot 1 - 3\alpha = -7 \implies 2 - 3\alpha = -7 \implies -3\alpha = -9 \implies \alpha = 3 \] \[ a = 2 \cdot 3 = 6 \] Check if $x + 2y = 0$ is common: \[ ax^2 - 7xy - 3y^2 = (x + 2y)(l'x + m'y) \] \[ = l'x^2 + (2l' + m')xy + 2m'y^2 \] \[ l' = a, \quad 2l' + m' = -7, \quad 2m' = -3 \implies m' = -\frac{3}{2} \] \[ 2a - \frac{3}{2} = -7 \implies 2a = -\frac{11}{2} \implies a = -\frac{11}{4} \] Since $a$ is an integer, $a = 6$. Thus, the first pair is: \[ 6x^2 - 7xy - 3y^2 = (2x - 3y)(3x + y) = 0 \] For a pair of lines $ax^2 + 2hxy + by^2 = 0$, the angle bisectors are: \[ \frac{x^2 - y^2}{a - b} = \frac{xy}{h} \] Here, $a = 6$, $b = -3$, $h = -\frac{7}{2}$: \[ \frac{x^2 - y^2}{6 - (-3)} = \frac{xy}{-\frac{7}{2}} \implies \frac{x^2 - y^2}{9} = -\frac{2xy}{7} \] \[ 7 (x^2 - y^2) = -18xy \implies 7x^2 + 18xy - 7y^2 = 0 \] Option (1) is correct. Options (2), (3), and (4) do not match the bisector equation.