At the intersection points of the line y = 0

At the intersection points of the line $y = 0$ and the parabola $y = x^2 - 4$, normals are drawn to the parabola. Find the area bounded by the parabola and the normals.

We will find the normal lines at the roots of the parabola and integrate the enclosed region using symmetry.

Step 1: Find intersection points

The parabola intersects the $x$-axis ($y = 0$) where:

\[0 = x^2 - 4\]

\[x^2 = 4\]

\[x = \pm 2\]

The intersection points are $(-2, 0)$ and $(2, 0)$.

Step 2: Find the normal lines

For the parabola $y = x^2 - 4$:

\[\frac{dy}{dx} = 2x\]

The slope of the tangent at point $(x_0, y_0)$ is $m_t = 2x_0$. The slope of the normal is:

\[m_n = -\frac{1}{m_t} = -\frac{1}{2x_0}\]

Normal at $(-2, 0)$: Slope: $m_1 = -\frac{1}{2(-2)} = \frac{1}{4}$

Equation: $y - 0 = \frac{1}{4}(x - (-2))$

\[y = \frac{1}{4}(x + 2) = \frac{x}{4} + \frac{1}{2}\]

Normal at $(2, 0)$: Slope: $m_2 = -\frac{1}{2(2)} = -\frac{1}{4}$

Equation: $y - 0 = -\frac{1}{4}(x - 2)$

\[y = -\frac{1}{4}(x - 2) = -\frac{x}{4} + \frac{1}{2}\]

Step 3: Find intersection of the two normals

Setting $\frac{x}{4} + \frac{1}{2} = -\frac{x}{4} + \frac{1}{2}$:

\[\frac{x}{4} = -\frac{x}{4}\]

\[\frac{x}{2} = 0\]

\[x = 0\]

\[y = \frac{0}{4} + \frac{1}{2} = \frac{1}{2}\]

The normals intersect at $\left(0, \frac{1}{2}\right)$.

Step 4: Determine the region

The region is bounded by:

• The parabola $y = x^2 - 4$ (from $x = -2$ to $x = 2$)
• The normal $y = \frac{x}{4} + \frac{1}{2}$ (from $x = -2$ to $x = 0$)
• The normal $y = -\frac{x}{4} + \frac{1}{2}$ (from $x = 0$ to $x = 2$)

The normals are above the parabola in this region (since the parabola opens upward and passes through the points where the normals start).

Step 5: Set up the integral

\[A = \int_{-2}^{0} \left[\left(\frac{x}{4} + \frac{1}{2}\right) - (x^2 - 4)\right] dx + \int_{0}^{2} \left[\left(-\frac{x}{4} + \frac{1}{2}\right) - (x^2 - 4)\right] dx\]

Simplifying the integrands:

First integral:

\[\frac{x}{4} + \frac{1}{2} - x^2 + 4 = -x^2 + \frac{x}{4} + \frac{9}{2}\]

Second integral:

\[-\frac{x}{4} + \frac{1}{2} - x^2 + 4 = -x^2 - \frac{x}{4} + \frac{9}{2}\]

Step 6: Use symmetry

Notice that the region is symmetric about the $y$-axis. We can compute:

\[A = 2\int_{0}^{2} \left(-x^2 - \frac{x}{4} + \frac{9}{2}\right) dx\]

Step 7: Evaluate the integral

\[A = 2\int_{0}^{2} \left(-x^2 - \frac{x}{4} + \frac{9}{2}\right) dx\]

\[= 2\left[-\frac{x^3}{3} - \frac{x^2}{8} + \frac{9x}{2}\right]_{0}^{2}\]

At $x = 2$:

\[-\frac{8}{3} - \frac{4}{8} + \frac{18}{2} = -\frac{8}{3} - \frac{1}{2} + 9\]

Converting to common denominator (6):

\[= -\frac{16}{6} - \frac{3}{6} + \frac{54}{6} = \frac{-16 - 3 + 54}{6} = \frac{35}{6}\]

At $x = 0$: $0$

Therefore:

\[A = 2 \cdot \frac{35}{6} = \frac{70}{6} = \frac{35}{3}\]

\[\boxed{A = \frac{35}{3} \text{ square units}}\]

Step 8: Numerical value

\[A = \frac{35}{3} \approx 11.667 \text{ square units}\]

Verification

Let's verify the calculation at $x = 2$:

\[-\frac{8}{3} - \frac{1}{2} + 9 = -\frac{16}{6} - \frac{3}{6} + \frac{54}{6} = \frac{35}{6} \quad \checkmark\]

Summary

The area bounded by the parabola $y = x^2 - 4$ and its normals at the points $(-2, 0)$ and $(2, 0)$ is:

\[\boxed{A = \frac{35}{3} \approx 11.667 \text{ square units}}\]

Key steps:

1. Find intersection points: $x^2 - 4 = 0 \implies x = \pm 2$
2. Find normal lines: $y = \frac{x}{4} + \frac{1}{2}$ and $y = -\frac{x}{4} + \frac{1}{2}$
3. Normals intersect at $\left(0, \frac{1}{2}\right)$
4. Use symmetry: $A = 2\int_{0}^{2} \left(-x^2 - \frac{x}{4} + \frac{9}{2}\right) dx$
5. Evaluate: $2 \cdot \frac{35}{6} = \frac{35}{3}$

Key insight:

The normal to a parabola at a point has slope perpendicular to the tangent. The region is symmetric about the $y$-axis, allowing us to simplify the calculation by doubling the area from one side.