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

Question

Aicharya · Accepted Answer

Given equation of curve is $\begin{aligned} & 3 x^2+(4-2 y) x+1-y^2=0 \\ & \therefore x=\frac{(2 y-4) \pm \sqrt{(4-2 y)^2-4 \times 3\left(1-y^2\right)}}{6} \\ & -\frac{2(y-2) \pm \sqrt{4(2 y-1)^2}}{6}=\frac{(y-2) \pm(2 y-1)}{3} \end{aligned}$ $\therefore$ Equation of pair of lines are $\begin{aligned} & L_1: x-y+1=0 ...(i)\\ & L_2: 3 x+y+1=0...(ii) \end{aligned}$
On solving (i) ằnd (ii), we get $x=\frac{-1}{2}$ and $y=\frac{1}{2}$
Another equation of curve is $\begin{aligned} & 3 x^2+(6-2 y) x+2 y-y^2=0 \\ & \Rightarrow x=\frac{(2 y-6) \pm \sqrt{(6-2 y)^2-4 \cdot 3 \cdot\left(2 y-y^2\right)}}{6} \\ & =\frac{(2 y-6) \pm 2(2 y-3)}{6}=\frac{(y-3) \pm(2 y-3)}{3} \end{aligned}$ $\therefore$ Equation of pair of lines are $\begin{aligned} & L_3: x-y+2=0...(iii) \\ & L_4: 3 x+y=0...(iv) \end{aligned}$
On solving (iii) and (iv) we get $x=\frac{-1}{2}$ and $y=\frac{3}{2}$

On solving (i) and (iv) we get $x=\frac{-1}{4}$ and $y=\frac{3}{4}$ On solving (ii) and (iii), we get $x=\frac{-3}{4}$ and $y=\frac{5}{4}$ $A C=\sqrt{0+\left(\frac{3}{2}-\frac{1}{2}\right)^2}=1$
Area of parallelogram $=\frac{1}{2} \times \frac{1}{4} \times 1+\frac{1}{2} \times \frac{1}{4} \times 1=\frac{1}{4}$.

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

Solution