If the bisectors of the angles of the lines r | vedclass

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(D) The equation of the pair of angle bisectors for a homogeneous equation $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h

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$90$

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(D) The equation of the pair of angle bisectors for a homogeneous equation $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.For the first equation $3x^2 - 4xy + 5y^2 = 0$,$a=3, h=-2, b=5$. The bisector equation is $\frac{x^2 - y^2}{3 - 5} = \frac{xy}{-2}$,which simplifies to $x^2 - y^2 = xy$,or $x^2 - xy - y^2 = 0$.For the second equation $5x^2 + 4xy + 3y^2 = 0$,$a=5, h=2, b=3$. The bisector equation is $\frac{x^2 - y^2}{5 - 3} = \frac{xy}{2}$,which simplifies to $x^2 - y^2 = xy$,or $x^2 - xy - y^2 = 0$.Since the bisectors are the same,the lines are related by a specific symmetry. The angle $	heta$ between the lines $a_1x^2 + 2h_1xy + b_1y^2 = 0$ and $a_2x^2 + 2h_2xy + b_2y^2 = 0$ is determined by their slopes. Given the symmetry and the bisector condition,the lines represented by the first equation are perpendicular to the lines represented by the second equation. Thus,the angle is $90^\circ$.

If the bisectors of the angles of the lines represented by $3x^2 - 4xy + 5y^2 = 0$ and $5x^2 + 4xy + 3y^2 = 0$ are the same,then the angle made by the lines represented by the first equation with the second is .....$^o$.

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