Question

The straight lines $7x^2+6xy+4y^2=0$ have the same pair of bisectors as those of the lines given by

• A. $49x^2+66xy+16y^2=0$
• B. $10x^2+6xy+7y^2=0$
• C. $5x^2+6xy+2y^2=0$
• D. $4x^2-6xy+7y^2=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $49x^2+66xy+16y^2=0$
B $10x^2+6xy+7y^2=0$
C $5x^2+6xy+2y^2=0$
D $4x^2-6xy+7y^2=0$

For $7x^2+6xy+4y^2=0$
Equation of angle bisector is
$\frac{x^2-y^2}{7-4}=\frac{xy}{3}\Rightarrow x^2-xy-y^2=0$
For Option A
Equation of angle bisector of $49x^2+66xy+16y^2=0$ is
$\frac{x^2-y^2}{33}=\frac{xy}{33}\Rightarrow x^2-xy-y^2=0$
For Option B
Equation of angle bisector of $10x^2+6xy+7y^2=0$ is
$\frac{x^2-y^2}{3}=\frac{xy}{3}\Rightarrow x^2-xy-y^2=0$
For Option C
Equation of angle bisector of $5x^2+6xy+2y^2=0$ is
$\frac{x^2-y^2}{3}=\frac{xy}{3}\Rightarrow x^2-xy-y^2=0$
For Option D
Equation of angle bisector of $4x^2-6xy+7y^2=0$ is
$\frac{x^2-y^2}{-3}=\frac{xy}{-3}\Rightarrow x^2-xy-y^2=0$