Area of the region enclosed by the curves 3x^ | vedclass

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(B) The given equations represent pairs of straight lines.For the first curve $3x^2-y^2-2xy+4x+1=0$,we can rewrite it as $3x^2+(4-2y)x+(1-y^2)=0$. Sol

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$\frac{1}{4}$

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(B) The given equations represent pairs of straight lines.For the first curve $3x^2-y^2-2xy+4x+1=0$,we can rewrite it as $3x^2+(4-2y)x+(1-y^2)=0$. Solving for $x$ using the quadratic formula:$x = \frac{-(4-2y) \pm \sqrt{(4-2y)^2 - 4(3)(1-y^2)}}{6} = \frac{(2y-4) \pm \sqrt{16-16y+4y^2-12+12y^2}}{6} = \frac{(2y-4) \pm \sqrt{16y^2-16y+4}}{6} = \frac{(2y-4) \pm 2(2y-1)}{6} = \frac{(y-2) \pm (2y-1)}{3}$.This gives two lines: $L_1: x-y+1=0$ and $L_2: 3x+y+1=0$.For the second curve $3x^2-y^2-2xy+6x+2y=0$,we rewrite it as $3x^2+(6-2y)x+(2y-y^2)=0$. Solving for $x$:$x = \frac{-(6-2y) \pm \sqrt{(6-2y)^2 - 4(3)(2y-y^2)}}{6} = \frac{(2y-6) \pm \sqrt{36-24y+4y^2-24y+12y^2}}{6} = \frac{(2y-6) \pm \sqrt{16y^2-48y+36}}{6} = \frac{(2y-6) \pm 2(2y-3)}{6} = \frac{(y-3) \pm (2y-3)}{3}$.This gives two lines: $L_3: x-y+2=0$ and $L_4: 3x+y=0$.The region is a parallelogram formed by the intersection of these four lines. The vertices are:$A = L_3 \cap L_4 = (-1/2, 3/2)$$B = L_1 \cap L_4 = (-1/4, 3/4)$$C = L_1 \cap L_2 = (-1/2, 1/2)$$D = L_2 \cap L_3 = (-3/4, 5/4)$The area of a parallelogram formed by lines $a_1x+b_1y+c_1=0, a_1x+b_1y+c_2=0, a_2x+b_2y+d_1=0, a_2x+b_2y+d_2=0$ is given by $\frac{|(c_1-c_2)(d_1-d_2)|}{|a_1b_2-a_2b_1|}$.Here,$L_1: x-y+1=0, L_3: x-y+2=0 \implies |c_1-c_2| = |1-2| = 1$.$L_2: 3x+y+1=0, L_4: 3x+y=0 \implies |d_1-d_2| = |1-0| = 1$.The denominator is $|(1)(1) - (3)(-1)| = |1+3| = 4$.Area $= \frac{1 	imes 1}{4} = \frac{1}{4}$.

Area of the region enclosed by the curves $3x^2-y^2-2xy+4x+1=0$ and $3x^2-y^2-2xy+6x+2y=0$ is

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