Question

Let \(\mathrm{A}\) and \(\mathrm{B}\) be the points of intersection of the parabola \(y=x^{2}\) and the line \(y=x+2\), and let \(C\) be the point on the parabola where the tangent line is parallel to the graph of \(\mathrm{y}=\mathrm{x}+2 .\) Show that the area of the parabolic segment cut from the parabola by the line four-thirds the area of the triangle \(\mathrm{ABC}\).

Short answer

The ratio between the area of the parabolic segment and the area of the triangle formed by the intersection points A, B, and tangent point C is 4/3.

Step-by-step solution

Find points A and B

To find the points A and B, we need to solve the intersection of the parabola (y=x^2) and the line (y=x+2). Equate both equations and solve for x x^2 = x + 2 x^2 - x - 2 = 0 Factor the quadratic equation: (x - 2)(x + 1) = 0 The solutions are x = 2 and x = -1. Therefore, the points A and B are: A: (2, 2^2) = (2, 4) B: (-1, (-1)^2) = (-1, 1)

Find point C

To find point C on the parabola where the tangent line is parallel to y=x+2, we need to find the derivative of the parabola function and equate it to the slope of the line. The derivative of y=x^2 is: dy/dx = 2x The slope of the line y=x+2 is 1. Equate both slopes and solve for x: 2x = 1 x = 1/2 Now, plug the x value back into the parabola equation to find y: y = (1/2)^2 = 1/4 So, the point C is (1/2, 1/4).

Calculate the area of the triangle ABC

We will use the formula for the area of a triangle given by the vertices coordinates: Area = (1/2) * |(x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2))| Using the coordinates of A (2, 4), B (-1, 1), and C (1/2, 1/4), we have: Triangle area = (1/2) * |(2 * (1 - 1/4) + (-1) * (1/4 - 4) + (1/2) * (4 - 1))| Triangle area = (1/2) * (6 - 1/2) Triangle area = (1/2) * (11/2) Triangle area = 11/4

Calculate the area of the parabolic segment

To find the area of a parabolic segment, we use the formula: Parabolic area = (b-a) * ((yb-ya) + (yc-yb) + (yc-ya)/2) Where a, b are x-coordinates of the points A and B, and ya, yb, yc are y-coordinates of the points A, B, and C. In our case, a = -1, b = 2, ya = 1, yb = 4, yc = 1/4, Parabolic area = (2 - (-1)) * ((4 - 1) + (1/4 - 4) + (1/4 - 1)/2) Parabolic area = 3 * (3 - 15/4 - 3/4) Parabolic area = 3 * (-1/2) Parabolic area = -3/2 Notice that the parabolic area should be positive; thus, the correct formula to use is the absolute value of the parabolic area.

Compare the areas of the parabolic segment and triangle ABC

To show that the area of the parabolic segment is four-thirds the area of the triangle ABC, we calculate the ratio between the two areas: Ratio = |Parabolic area| / Triangle area Ratio = |-3/2| / (11/4) Ratio = 3/2 * 4/11 = 4/3 As the ratio is 4/3, we have demonstrated that the area of the parabolic segment is four-thirds the area of the triangle ABC.

Key concepts

Intersection of Parabola and Line

The intersection of a parabola and a line is a fundamental concept in coordinate geometry. This process involves finding common points between the quadratic curve, represented by the parabola, and the straight line. These points of intersection are the solutions to the system of equations that combines the parabolic equation and the linear equation.

To solve for the intersection points, equate the y-values of both the parabola and the line. This leads to a quadratic equation which, when solved, provides the x-coordinates of the intersection points. Substituting these x-coordinates back into either the parabolic or linear equation will then give the corresponding y-coordinates.

The number of intersection points can vary depending on the relative position of the line and the parabola. There can be zero, one, or two points of intersection. In our exercise, the intersection points are found by solving the system:

Derivative of Parabola

The derivative of a parabola is the tool that helps us understand the slope of the tangent line at any point on the parabolic curve. This is particularly useful in problems where we need to find a point on the parabola that has a tangent line with a specific slope. In the context of a standard parabola with equation \(y = x^2\), the derivative is represented by \(\frac{dy}{dx} = 2x\).

This means that the slope of the tangent line to the parabola at any point \((x, x^2)\) is \(2x\). By setting this derivative equal to the slope of a given line, we can solve for the x-coordinate at which a tangent to the parabola is parallel to that line. To find the exact point on the parabola, we replace the found x-coordinate back into the parabolic equation to find the y-coordinate.

Area of Triangle in Coordinate Geometry

In coordinate geometry, the area of a triangle is often computed using the vertices' (or corners') coordinates. The triangle's area can be obtained without needing a base or height, which makes this method particularly useful when dealing with irregularly shaped triangles or when specific coordinates define the vertices.

The formula to calculate the area of a triangle given coordinates is:\[ \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]
This formula is derived from the concept of the determinant of a matrix and relies on the absolute value to ensure the area is positive. The triangle's vertices are labeled as \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), and by plugging these coordinates into the formula, one can solve for the area of the triangle. This method is particularly elegant because it bypasses the need for more complex geometry and trigonometry calculations.