Question:

Step 1: Identify the equations of the curves. The given equations are: Parabola: $y^2 = 4(x+1)$ Ellipse: $4x^2 + 9y^2 = 36$ Step 2: Find the intersection points. Substitute the expression for $y^2$ from the parabola equation into the ellipse equation: $$4x^2 + 9(4(x+1)) = 36$$ $$4x^2 + 36(x+1) = 36$$ $$4x^2 + 36x + 36 = 36$$ Subtract 36 from both sides: $$4x^2 + 36x = 0$$ Factor out $4x$: $$4x(x+9) = 0$$ This equation yields two possible values for $x$: $$x = 0 \quad \text{or} \quad x = -9$$ Step 3: Determine the corresponding y-coordinates for real intersection points. For $x=0$: Substitute $x=0$ into the parabola equation: $$y^2 = 4(0+1)$$ $$y^2 = 4$$ $$y = \pm 2$$ This gives two intersection points: $(0, 2)$ and $(0, -2)$. For $x=-9$: Substitute $x=-9$ into the parabola equation: $$y^2 = 4(-9+1)$$ $$y^2 = 4(-8)$$ $$y^2 = -32$$ Since $y^2$ cannot be negative for real numbers, there are no real intersection points when $x=-9$. Thus, the only real intersection points of the two curves are $(0, 2)$ and $(0, -2)$. Step 4: Calculate the length of the common chord. The common chord is the line segment connecting the intersection points $(0, 2)$ and $(0, -2)$. The distance between these two points is calculated using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d = \sqrt{(0 - 0)^2 + (-2 - 2)^2}$$ $$d = \sqrt{0^2 + (-4)^2}$$ $$d = \sqrt{0 + 16}$$ $$d = \sqrt{16}$$ $$d = 4$$ The length of the common chord is 4.