Solving A System Of Conic Equations: A Step-by-Step Guide
Hey guys! Today, we're going to dive into the fascinating world of conic sections and, more specifically, how to solve a system of conic equations. If you've ever wondered how circles and ellipses can intersect and what those intersection points actually are, you're in the right place. We'll be tackling a specific problem: finding the solutions to the system of equations (x-1)2+(y+4)2=25 and ((x-1)2)/25+((y+4)2)/100=1. Buckle up, because we're about to get our math on!
Understanding Conic Sections
Before we jump into the nitty-gritty of solving the system, let's take a moment to understand what conic sections are. Conic sections are curves formed by the intersection of a plane and a double right circular cone. The main types of conic sections are circles, ellipses, parabolas, and hyperbolas. Each of these shapes has its own unique equation and properties, which make them super interesting to study. Understanding their equations is the foundation for finding their intersection points when they form a system.
- Circles: A circle is the set of all points in a plane that are equidistant from a center point. The standard equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h, k) is the center and r is the radius. In our problem, the first equation, (x-1)2+(y+4)2=25, represents a circle with a center at (1, -4) and a radius of 5. Visualizing this circle can be the first step in understanding the problem. Imagine plotting this circle on a graph; it gives you a clear picture of the possible solutions. The circle's center and radius act as key parameters that define its size and position in the coordinate plane. Without a solid grasp of these parameters, solving problems related to circles, especially systems of equations involving circles, can be quite challenging.
- Ellipses: An ellipse is like a stretched-out circle. It's the set of all points such that the sum of the distances from two fixed points (called foci) is constant. The standard equation of an ellipse centered at (h, k) is ((x-h)2)/a2 + ((y-k)2)/b2 = 1, where 'a' and 'b' are the semi-major and semi-minor axes, respectively. The second equation in our system, ((x-1)2)/25+((y+4)2)/100=1, represents an ellipse also centered at (1, -4). The semi-major axis (along the y-axis) is 10, and the semi-minor axis (along the x-axis) is 5. When you're thinking about an ellipse, remember its stretched circular shape, and how the lengths of its axes determine its elongation. The interplay between these axes and the ellipse's center gives it a unique orientation and size in the plane. Grasping these characteristics makes it easier to understand how an ellipse might intersect with other conic sections, such as our circle.
Setting Up the System of Equations
Now that we've refreshed our knowledge of circles and ellipses, let's revisit our system of equations:
- (x-1)^2 + (y+4)^2 = 25
- ((x-1)^2)/25 + ((y+4)^2)/100 = 1
Notice that both equations have terms (x-1)^2 and (y+4)^2. This commonality is a golden opportunity for simplification using substitution. Recognizing such patterns is crucial in solving complex mathematical problems efficiently. The presence of these repeated terms not only simplifies our calculations but also provides a visual connection between the two conics, as they share the same center coordinates when considered in their standard forms. By spotting these shared elements early on, we can streamline our approach and sidestep unnecessary complications. This technique of recognizing and utilizing common structures is a hallmark of effective problem-solving in mathematics.
Solving by Substitution
Here's where the magic happens! Let's use substitution to solve this system. This method involves replacing a complex expression with a simpler variable, making the equations easier to manipulate. It's a bit like giving the complex terms a nickname so we can handle them more easily. Substitution is a powerful tool in algebra and calculus, often used to simplify integrals and differential equations as well. It works by temporarily changing the variables in an equation to make it more manageable, and it’s particularly useful when dealing with systems of equations that have common, repeated expressions.
Step 1: Introduce New Variables
To make things cleaner, let's substitute:
- A = (x-1)^2
- B = (y+4)^2
Our equations now look much friendlier:
- A + B = 25
- A/25 + B/100 = 1
Step 2: Solve for One Variable
From equation (1), we can easily express A in terms of B:
A = 25 - B
Step 3: Substitute into the Second Equation
Now, let's plug this value of A into equation (2):
(25 - B)/25 + B/100 = 1
Step 4: Simplify and Solve for B
To get rid of the fractions, multiply the entire equation by 100:
4(25 - B) + B = 100
100 - 4B + B = 100
-3B = 0
B = 0
Step 5: Find A
Now that we have B, we can find A using our expression from Step 2:
A = 25 - B
A = 25 - 0
A = 25
Back-Substitution to Find x and y
We've found A and B, but remember, we're ultimately looking for x and y. Time to back-substitute! This is where we reverse our initial substitution, replacing our stand-in variables with their original expressions to finally solve for x and y. Back-substitution is a fundamental technique in solving systems of equations, and it's not just limited to algebraic problems. You'll find this method incredibly useful in various areas of math, like differential equations and linear algebra, where you often introduce temporary variables to simplify calculations.
Step 1: Substitute Back
Recall our substitutions:
- A = (x-1)^2 = 25
- B = (y+4)^2 = 0
Step 2: Solve for x
Let's solve for x using A = (x-1)^2 = 25:
(x-1)^2 = 25
Taking the square root of both sides:
x - 1 = ±5
So, we have two possible values for x:
x = 1 + 5 = 6
x = 1 - 5 = -4
Step 3: Solve for y
Now, let's solve for y using B = (y+4)^2 = 0:
(y+4)^2 = 0
Taking the square root:
y + 4 = 0
y = -4
The Solutions
We've done it! We found the solutions to the system of conic equations. Our solutions are the points where the circle and ellipse intersect.
The Solutions are: (6, -4) and (-4, -4)
These are the points where the circle and ellipse touch in the coordinate plane. Visualizing these points on a graph alongside the circle and ellipse can provide an intuitive understanding of the solution. The solutions represent not just algebraic answers but also geometrical intersections, highlighting the beautiful interplay between algebra and geometry in mathematics.
Verification
It's always a good idea to verify our solutions to make sure they're correct. Plug the solutions back into the original equations to ensure they hold true. This step can help catch any arithmetic errors or missteps along the way, reinforcing the reliability of our results. Verifying solutions is a best practice in problem-solving across various mathematical disciplines, including calculus and differential equations.
Verify (6, -4):
- (6-1)^2 + (-4+4)^2 = 25 --> 25 + 0 = 25 (True)
- ((6-1)^2)/25 + ((-4+4)^2)/100 = 1 --> 25/25 + 0/100 = 1 (True)
Verify (-4, -4):
- (-4-1)^2 + (-4+4)^2 = 25 --> 25 + 0 = 25 (True)
- ((-4-1)^2)/25 + ((-4+4)^2)/100 = 1 --> 25/25 + 0/100 = 1 (True)
Both solutions check out! This step-by-step process of verification not only confirms the accuracy of our answers but also deepens our understanding of the problem and the relationships between the equations.
Conclusion
Solving systems of conic equations can seem daunting at first, but by breaking it down into smaller, manageable steps, it becomes much easier. We used substitution to simplify the equations, solved for the variables, and then back-substituted to find our x and y values. Remember, the key is to understand the properties of the conic sections involved and to use algebraic techniques wisely. Keep practicing, and you'll become a conic section master in no time! Whether it's circles, ellipses, or other conics, the methods we've explored here provide a solid foundation for tackling more complex problems in algebra and geometry.
I hope this guide has been helpful, guys! Happy solving!