If the angle between the pair of lines x^2+2 | vedclass

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(A) The given pair of lines is $x^2+2 \sqrt{2} x y+k y^2=0$. Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=\sqrt{2}, b=k$.The angle $	heta$ betwe

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$\frac{1}{3}$

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(A) The given pair of lines is $x^2+2 \sqrt{2} x y+k y^2=0$. Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=\sqrt{2}, b=k$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$.Given $	heta = 45^{\circ}$,so $	an 45^{\circ} = 1 = \left| \frac{2\sqrt{2-k}}{1+k} ight|$.Squaring both sides: $1 = \frac{4(2-k)}{(1+k)^2}$ $\Rightarrow (1+k)^2 = 8-4k$ $\Rightarrow 1+k^2+2k = 8-4k$ $\Rightarrow k^2+6k-7=0$.Solving for $k$: $(k+7)(k-1)=0$. Since $k>0$,we get $k=1$.The pair of lines is $x^2+2\sqrt{2}xy+y^2=0$. The equation of the angle bisectors is $\frac{x^2-y^2}{a-b} = \frac{xy}{h} \Rightarrow \frac{x^2-y^2}{1-1} = \frac{xy}{\sqrt{2}}$,which is not the standard form. Using $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$ is invalid when $a=b$. The correct bisector equation is $h(x^2-y^2) = (a-b)xy$. Since $a=b=1$,we have $0 = \sqrt{2}(x^2-y^2)$,so $x^2-y^2=0$,which gives $x-y=0$ and $x+y=0$.The triangle is formed by $x-y=0, x+y=0$ and $x+2y+1=0$. The vertices are $(0,0)$,$(-1,0)$,and $(-1/3, -1/3)$.The area is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = \frac{1}{2} |0 + (-1)(-1/3-0) + (-1/3)(0-0)| = \frac{1}{2} |1/3| = \frac{1}{6}$.

If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2=0, k>0$ is $45^{\circ}$,then the area (in square units) of the triangle formed by the pair of bisectors of the angles between these lines and the line $x+2 y+1=0$ is

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