Finding Parabola Equation Passing Through Points (1,1), (2,2), And (-1,5)
In the realm of mathematics, particularly in algebra and analytic geometry, parabolas hold a significant position. These U-shaped curves, defined as the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix), appear in various real-world applications, from the trajectory of a projectile to the design of satellite dishes. One common problem encountered in mathematics involves determining the equation of a parabola that passes through a given set of points. This article delves into a specific instance of this problem, where we aim to find the equation of a parabola that passes through the points (1,1), (2,2), and (-1,5), with the added constraint of a vertical axis of symmetry.
Before we dive into the solution, it's crucial to have a solid understanding of parabolas and their equations. A parabola with a vertical axis of symmetry can be represented by the general quadratic equation:
y = ax^2 + bx + c
where a, b, and c are constants that determine the shape and position of the parabola. The axis of symmetry is a vertical line that passes through the vertex of the parabola, dividing it into two symmetrical halves. The x-coordinate of the vertex is given by -b/(2a). The constraint of a vertical axis of symmetry simplifies our problem, as it restricts the possible orientations of the parabola.
To find the specific equation of the parabola that passes through the given points, we need to determine the values of the constants a, b, and c. Since we are given three points, we can substitute their coordinates into the general equation to obtain three equations with three unknowns. This system of equations can then be solved to find the values of a, b, and c, thus defining the equation of the parabola.
We are given the points (1,1), (2,2), and (-1,5). Substituting these points into the general equation of the parabola, we get the following system of equations:
- For (1,1): 1 = a(1)^2 + b(1) + c => 1 = a + b + c
- For (2,2): 2 = a(2)^2 + b(2) + c => 2 = 4a + 2b + c
- For (-1,5): 5 = a(-1)^2 + b(-1) + c => 5 = a - b + c
Now we have a system of three linear equations with three unknowns: a, b, and c. We can use various methods to solve this system, such as substitution, elimination, or matrix methods.
Let's use the elimination method to solve the system of equations. We can start by subtracting equation (1) from equation (2) and equation (3) to eliminate c:
- Equation (2) - Equation (1): (4a + 2b + c) - (a + b + c) = 2 - 1 => 3a + b = 1 (4)
- Equation (3) - Equation (1): (a - b + c) - (a + b + c) = 5 - 1 => -2b = 4
From the second equation, we can easily solve for b:
- -2b = 4 => b = -2
Now that we have the value of b, we can substitute it into equation (4) to solve for a:
- 3a + (-2) = 1 => 3a = 3 => a = 1
Finally, we can substitute the values of a and b into equation (1) to solve for c:
- 1 = 1 + (-2) + c => 1 = -1 + c => c = 2
Therefore, we have found the values of the constants: a = 1, b = -2, and c = 2.
Now that we have the values of a, b, and c, we can write the equation of the parabola:
y = x^2 - 2x + 2
This is the equation of the parabola that passes through the points (1,1), (2,2), and (-1,5) and has a vertical axis of symmetry.
To ensure that our solution is correct, we can substitute the coordinates of the given points back into the equation and verify that they satisfy the equation:
- For (1,1): 1 = (1)^2 - 2(1) + 2 => 1 = 1 - 2 + 2 => 1 = 1 (Correct)
- For (2,2): 2 = (2)^2 - 2(2) + 2 => 2 = 4 - 4 + 2 => 2 = 2 (Correct)
- For (-1,5): 5 = (-1)^2 - 2(-1) + 2 => 5 = 1 + 2 + 2 => 5 = 5 (Correct)
Since all three points satisfy the equation, we can confidently conclude that our solution is correct.
Now that we have the equation of the parabola, we can determine some of its key properties. The equation is:
y = x^2 - 2x + 2
Vertex
The vertex of the parabola is the point where the parabola changes direction. The x-coordinate of the vertex is given by -b/(2a). In this case, a = 1 and b = -2, so the x-coordinate of the vertex is:
x = -(-2) / (2 * 1) = 1
To find the y-coordinate of the vertex, we substitute x = 1 into the equation of the parabola:
y = (1)^2 - 2(1) + 2 = 1 - 2 + 2 = 1
Therefore, the vertex of the parabola is (1,1).
Axis of Symmetry
The axis of symmetry is a vertical line that passes through the vertex of the parabola. Since the x-coordinate of the vertex is 1, the equation of the axis of symmetry is:
x = 1
Y-intercept
The y-intercept is the point where the parabola intersects the y-axis. To find the y-intercept, we set x = 0 in the equation of the parabola:
y = (0)^2 - 2(0) + 2 = 2
Therefore, the y-intercept is (0,2).
Concavity
The concavity of the parabola is determined by the sign of the coefficient a. In this case, a = 1, which is positive. Therefore, the parabola opens upwards and is concave up.
Graphing the Parabola
To graph the parabola, we can plot the vertex, the y-intercept, and the given points. We can also find additional points by substituting different values of x into the equation. The graph of the parabola will be a U-shaped curve that is symmetrical about the axis of symmetry.
While we used the elimination method to solve the system of equations, there are other methods that could be used, such as:
- Substitution Method: Solve one equation for one variable and substitute that expression into the other equations.
- Matrix Methods: Represent the system of equations as a matrix equation and solve using matrix operations, such as Gaussian elimination or matrix inversion.
Each method has its advantages and disadvantages, and the choice of method often depends on the specific system of equations and personal preference.
The method we used to find the equation of a parabola passing through specific points can be generalized to find the equation of any conic section (parabola, ellipse, hyperbola) that passes through a given set of points. The general approach involves setting up a system of equations based on the general equation of the conic section and the given points, and then solving the system to find the coefficients in the equation.
Parabolas have numerous applications in various fields, including:
- Physics: The trajectory of a projectile under the influence of gravity is a parabola.
- Engineering: Parabolic mirrors and reflectors are used in telescopes, satellite dishes, and solar collectors.
- Architecture: Parabolic arches are used in bridges and buildings.
- Mathematics: Parabolas are fundamental in calculus, analytic geometry, and other areas of mathematics.
In this article, we have demonstrated how to find the equation of a parabola that passes through the points (1,1), (2,2), and (-1,5) with a vertical axis of symmetry. We began by understanding the general equation of a parabola with a vertical axis of symmetry and then set up a system of equations by substituting the coordinates of the given points into the general equation. We solved this system using the elimination method to find the values of the constants a, b, and c, which allowed us to write the specific equation of the parabola. We then verified our solution by substituting the given points back into the equation. Additionally, we explored key properties of the parabola, such as its vertex, axis of symmetry, y-intercept, and concavity. We also discussed alternative methods for solving the system of equations and highlighted the broader applications of parabolas in various fields.
This problem showcases the power of algebraic techniques in solving geometric problems. By translating geometric conditions into algebraic equations, we can leverage the tools of algebra to find solutions and gain insights into the properties of geometric shapes. The ability to find the equation of a parabola that passes through specific points is a valuable skill in mathematics and has practical applications in various scientific and engineering disciplines. Understanding parabolas and their properties is crucial for success in many areas of mathematics and its applications.
