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Magazine Chapter 23: Problem 46
The area bounded by the parabola \(y^{2}=4 x\) and the line \(2 x-3 y+4=0\), in square unit, is \(\quad\) [Online May 26, 2012] (a) \(\frac{2}{5}\) (b) \(\frac{1}{3}\) (c) 1 (d) \(\frac{1}{2}\)
Short Answer
Expert verified
The area is \( \frac{1}{3} \) square units (option b).
Step by step solution
01
Identify the curves
The given curves are a parabola \( y^2 = 4x \) and a line \( 2x - 3y + 4 = 0 \). We need to express these in standard forms. The parabola is standard as is, but the line can be rearranged to \( x = \frac{3y - 4}{2} \).
02
Find intersection points
To find the intersection points, substitute \( x = \frac{3y - 4}{2} \) into the parabola equation. So, \( y^2 = 4 \left( \frac{3y - 4}{2} \right) \). Simplifying gives \( y^2 = 6y - 8 \) or \( y^2 - 6y + 8 = 0 \).
03
Solve quadratic equation
Solve the quadratic equation \( y^2 - 6y + 8 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = -6, c = 8 \). The solutions will give us the y-coordinates of the intersections.
04
Calculate roots
Calculate the discriminant \( b^2 - 4ac = (-6)^2 - 4(1)(8) = 36 - 32 = 4 \). So, \( y = \frac{6 \pm \sqrt{4}}{2} = \frac{6 \pm 2}{2} \). The roots are \( y = 4 \) and \( y = 2 \).
05
Find corresponding x-coordinates
Substitute \( y = 4 \) and \( y = 2 \) back into \( x = \frac{3y - 4}{2} \) to find corresponding x-coordinates. For \( y = 4 \), \( x = \frac{3(4) - 4}{2} = \frac{12 - 4}{2} = 4 \). For \( y = 2 \), \( x = \frac{3(2) - 4}{2} = \frac{6 - 4}{2} = 1 \). So, intersections: \((4,4)\) and \((1,2)\).
06
Set up area integral
The area between the parabola and the line from \( y = 2 \) to \( y = 4 \) is we integrate the difference in x-coordinates, \( x_{line} - x_{parabola} \). Evaluate integral of \( \frac{3y-4}{2} - \frac{y^2}{4} \) with respect to y, from 2 to 4.
07
Compute the integral
Integrate \( \frac{3y-4}{2} - \frac{y^2}{4} \) between \( y = 2 \) and \( y = 4 \): \( \int_{2}^{4} \left( \frac{3y-4}{2} - \frac{y^2}{4} \right) \, dy \). Compute the antiderivative and evaluate at the bounds.
08
Evaluate definite integral
Find the antiderivative: \( \frac{3}{2} \cdot 1/2 y^2 - \frac{4}{2} y - \frac{1}{4} \cdot 1/3 y^3 \). Evaluating from 2 to 4 gives \( \left[ \frac{3}{4}y^2 - 2y - \frac{1}{12}y^3 \right]_2^4 \). Substitute and calculate the result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
A parabola is a symmetrical, U-shaped curve described by a quadratic equation. In this exercise, the parabolic equation is given as \( y^2 = 4x \). This form is a specific type of parabola that opens horizontally.
Parabolas have certain key features:
Parabolas have certain key features:
- The vertex is the turning point, which in this case is at the origin (0,0).
- This parabola opens to the right, as indicated by the positive coefficient of \( x \).
- The directrix, which is a line perpendicular to the axis of symmetry, is \( x = -1 \), at a distance \( 1 \) unit opposite to the coefficient of \( x \).
- This parabola is symmetric about the x-axis.
Intersection points
Finding intersection points is a critical step in determining the area between two curves. Intersection points occur where the two curves share the same values of \( x \) and \( y \). In our exercise, we used algebraic substitution:
- Substitute the expression for \( x \) from the line equation \( x = \frac{3y - 4}{2} \) into the parabola's equation \( y^2 = 4x \).
- This substitution transforms the problem into a single-variable quadratic equation, \( y^2 - 6y + 8 = 0 \).
- Solving this equation yields the \( y \)-coordinates of the intersection points, in this case, \( y = 4 \) and \( y = 2 \).
- Plug these \( y \) values into the line equation to find \( x \)-coordinates: \((4,4)\) and \((1,2)\).
Integral calculus
Integral calculus is essential for finding the area between curves. It involves calculating the integral, which is the accumulation of quantities over a region defined by the curves. In this problem, we calculate the definite integral:
- Set up the integral to measure the 'area under the curve' between the intersection points.
- Calculate the integral of the difference \( \left( \frac{3y-4}{2} - \frac{y^2}{4} \right) \) over the y-range from \( 2 \) to \( 4 \).
- Evaluate the integral to find the net area between the curves over this interval.
Quadratic equation
A quadratic equation is characterized by the form \( ax^2 + bx + c = 0 \), and solving it helps find critical points such as intersections.
In our problem, we derived the quadratic equation \( y^2 - 6y + 8 = 0 \) when finding the intersection points:
In our problem, we derived the quadratic equation \( y^2 - 6y + 8 = 0 \) when finding the intersection points:
- Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( y \).
- Calculate the discriminant \( b^2 - 4ac \) to determine the nature of the roots: a positive discriminant indicates two distinct real roots.
- The solutions are \( y = 4 \) and \( y = 2 \), providing critical interaction points for calculating the bounded area.
