Question:

Step 1: Factorize the given equation to identify the constituent lines. The given equation is $x^2y-y^3-x^2+5y^2-8y+4=0$. We can group terms involving $x^2$: $$x^2(y-1) - y^3 + 5y^2 - 8y + 4 = 0$$ Now, consider the polynomial in $y$: $P(y) = -y^3 + 5y^2 - 8y + 4$. We test for integer roots. For $y=1$: $$P(1) = -(1)^3 + 5(1)^2 - 8(1) + 4 = -1 + 5 - 8 + 4 = 0$$ Since $P(1)=0$, $(y-1)$ is a factor of $P(y)$. We can perform polynomial division or factor by grouping: $$-y^3 + 5y^2 - 8y + 4 = -y^2(y-1) + 4y^2 - 8y + 4$$ $$= -y^2(y-1) + 4(y^2 - 2y + 1)$$ $$= -y^2(y-1) + 4(y-1)^2$$ $$= (y-1)(-y^2 + 4(y-1))$$ $$= (y-1)(-y^2 + 4y - 4)$$ $$= -(y-1)(y^2 - 4y + 4)$$ $$= -(y-1)(y-2)^2$$ Substitute this back into the main equation: $$x^2(y-1) - (y-1)(y-2)^2 = 0$$ Factor out $(y-1)$: $$(y-1)[x^2 - (y-2)^2] = 0$$ Step 2: Identify the equations of the straight lines. The factored equation $(y-1)[x^2 - (y-2)^2] = 0$ implies that either $(y-1)=0$ or $[x^2 - (y-2)^2]=0$. From $(y-1)=0$, we get the first line: $$L_1: y=1$$ From $[x^2 - (y-2)^2]=0$, we can use the difference of squares formula $a^2-b^2=(a-b)(a+b)$: $$(x - (y-2))(x + (y-2)) = 0$$ This gives two more lines: $$(x - y + 2) = 0 \implies y = x+2 \quad (L_2)$$ $$(x + y - 2) = 0 \implies y = -x+2 \quad (L_3)$$ Thus, the given equation represents three straight lines: $y=1$, $y=x+2$, and $y=-x+2$. Step 3: Determine the vertices of the region bounded by these lines. The region is a triangle formed by the intersection of these three lines. 1. Intersection of $L_1$ and $L_2$: $y=1$ and $y=x+2$. Substituting $y=1$ into $y=x+2$: $1 = x+2 \implies x = -1$. Vertex $A = (-1, 1)$. 2. Intersection of $L_1$ and $L_3$: $y=1$ and $y=-x+2$. Substituting $y=1$ into $y=-x+2$: $1 = -x+2 \implies x = 1$. Vertex $B = (1, 1)$. 3. Intersection of $L_2$ and $L_3$: $y=x+2$ and $y=-x+2$. Equating the expressions for $y$: $x+2 = -x+2$ $2x = 0 \implies x=0$. Substituting $x=0$ into $y=x+2$: $y=0+2 \implies y=2$. Vertex $C = (0, 2)$. The vertices of the triangle are $A(-1, 1)$, $B(1, 1)$, and $C(0, 2)$. Step 4: Calculate the area of the triangle. We can use the base-height formula for the area of a triangle. Consider the segment $AB$ as the base. $A(-1, 1)$ and $B(1, 1)$ lie on the line $y=1$. The length of the base $AB$ is the distance between $x=-1$ and $x=1$ along $y=1$: $$ \text{Base} = |1 - (-1)| = |2| = 2 \text{ units} $$ The height of the triangle is the perpendicular distance from vertex $C(0, 2)$ to the base line $y=1$. The y-coordinate of $C$ is 2, and the y-coordinate of the base line is 1. $$ \text{Height} = |2 - 1| = 1 \text{ unit} $$ The area of the triangle is given by: $$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} $$ $$ \text{Area} = \frac{1}{2} \times 2 \times 1 = 1 \text{ square unit} $$ The final answer is $\boxed{\text{1}}$.