At the intersection points of the line x - y + 1 = 0

At the intersection points of the line $x - y + 1 = 0$ and the parabola $y = x^2 - 4x + 5$, tangents are drawn to the parabola. Calculate the area bounded by the parabola and the tangents.

We will find the tangent lines at the intersection points and integrate the region they enclose with the parabola.

Step 1: Find intersection points

The line is: $y = x + 1$. Setting equal to the parabola:

\[x + 1 = x^2 - 4x + 5\]

\[x^2 - 5x + 4 = 0\]

\[(x - 4)(x - 1) = 0\]

So $x = 1$ or $x = 4$.

Point 1: $x = 1 \implies y = 1 + 1 = 2$, so $(1, 2)$

Point 2: $x = 4 \implies y = 4 + 1 = 5$, so $(4, 5)$

Step 2: Find the tangent lines

The parabola is $y = x^2 - 4x + 5$. The derivative is:

\[\frac{dy}{dx} = 2x - 4\]

Tangent at $(1, 2)$: Slope: $m_1 = 2(1) - 4 = -2$

Equation: $y - 2 = -2(x - 1)$

\[y = -2x + 2 + 2 = -2x + 4\]

Tangent at $(4, 5)$: Slope: $m_2 = 2(4) - 4 = 4$

Equation: $y - 5 = 4(x - 4)$

\[y = 4x - 16 + 5 = 4x - 11\]

Step 3: Find intersection of the two tangents

Setting $-2x + 4 = 4x - 11$:

\[-2x + 4 = 4x - 11\]

\[15 = 6x\]

\[x = \frac{5}{2}\]

\[y = -2 \cdot \frac{5}{2} + 4 = -5 + 4 = -1\]

The tangents intersect at $\left(\frac{5}{2}, -1\right)$.

Step 4: Set up the region

The region is bounded by:

• The parabola $y = x^2 - 4x + 5$ (from $x = 1$ to $x = 4$)
• The tangent $y = -2x + 4$ (from $x = 1$ to $x = \frac{5}{2}$)
• The tangent $y = 4x - 11$ (from $x = \frac{5}{2}$ to $x = 4$)

The parabola is above the tangents (since it's convex upward).

Step 5: Set up the integral

\[A = \int_1^{5/2} \left[(x^2 - 4x + 5) - (-2x + 4)\right] dx + \int_{5/2}^4 \left[(x^2 - 4x + 5) - (4x - 11)\right] dx\]

Simplifying:

First integral:

\[(x^2 - 4x + 5) - (-2x + 4) = x^2 - 4x + 5 + 2x - 4 = x^2 - 2x + 1 = (x - 1)^2\]

Second integral:

\[(x^2 - 4x + 5) - (4x - 11) = x^2 - 4x + 5 - 4x + 11 = x^2 - 8x + 16 = (x - 4)^2\]

Therefore:

\[A = \int_1^{5/2} (x - 1)^2 \, dx + \int_{5/2}^4 (x - 4)^2 \, dx\]

Step 6: Evaluate the integrals

First integral:

\[\int_1^{5/2} (x - 1)^2 \, dx = \left[\frac{(x - 1)^3}{3}\right]_1^{5/2}\]

At $x = \frac{5}{2}$:

\[\frac{\left(\frac{5}{2} - 1\right)^3}{3} = \frac{\left(\frac{3}{2}\right)^3}{3} = \frac{27/8}{3} = \frac{27}{24} = \frac{9}{8}\]

At $x = 1$: $\frac{(1 - 1)^3}{3} = 0$

So: $\int_1^{5/2} (x - 1)^2 \, dx = \frac{9}{8}$

Second integral:

\[\int_{5/2}^4 (x - 4)^2 \, dx = \left[\frac{(x - 4)^3}{3}\right]_{5/2}^4\]

At $x = 4$: $\frac{(4 - 4)^3}{3} = 0$

At $x = \frac{5}{2}$:

\[\frac{\left(\frac{5}{2} - 4\right)^3}{3} = \frac{\left(-\frac{3}{2}\right)^3}{3} = \frac{-27/8}{3} = -\frac{27}{24} = -\frac{9}{8}\]

So: $\int_{5/2}^4 (x - 4)^2 \, dx = 0 - \left(-\frac{9}{8}\right) = \frac{9}{8}$

Step 7: Total area

\[A = \frac{9}{8} + \frac{9}{8} = \frac{18}{8} = \frac{9}{4}\]

\[\boxed{A = \frac{9}{4} \text{ square units}}\]

Step 8: Numerical value

\[A = \frac{9}{4} = 2.25 \text{ square units}\]

Summary

The area bounded by the parabola $y = x^2 - 4x + 5$ and its tangents at the intersection points with the line $x - y + 1 = 0$ is:

\[\boxed{A = \frac{9}{4} = 2.25 \text{ square units}}\]

Key steps:

1. Find intersection points: $(1, 2)$ and $(4, 5)$
2. Find tangent lines: $y = -2x + 4$ and $y = 4x - 11$
3. Tangents intersect at $\left(\frac{5}{2}, -1\right)$
4. Set up integrals: $\int_1^{5/2} (x-1)^2 dx + \int_{5/2}^4 (x-4)^2 dx$
5. Evaluate: $\frac{9}{8} + \frac{9}{8} = \frac{9}{4}$

Key insight:

The region naturally splits at the intersection of the tangents, and by symmetry of the squared expressions $(x-1)^2$ and $(x-4)^2$, both integrals contribute equally to the total area.