Find The Center Of An Ellipse X^2+4y^2-10x-40y+121=0

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Finding the center of an ellipse can seem daunting at first, but with a systematic approach, it becomes a manageable task. This article provides a comprehensive, step-by-step guide on how to determine the center of an ellipse given its equation. We will focus on the specific equation: x2+4y2−10x−40y+121=0x^2 + 4y^2 - 10x - 40y + 121 = 0. By understanding the underlying principles and applying the method of completing the square, you'll be able to confidently find the center of any ellipse.

Understanding the Ellipse Equation

Before diving into the solution, let's understand the general form of an ellipse equation. An ellipse is essentially a stretched circle, and its equation reflects this geometric characteristic. The standard form of an ellipse equation centered at point (h,k)(h, k) is:

(x−h)2a2+(y−k)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1

Where:

  • (h,k)(h, k) represents the coordinates of the center of the ellipse.
  • aa is the semi-major axis (the distance from the center to the ellipse along the horizontal axis if a>ba > b or along the vertical axis if a<ba < b).
  • bb is the semi-minor axis (the distance from the center to the ellipse along the other axis).

Sometimes, the equation is given in a general form, which looks like this:

Ax2+Cy2+Dx+Ey+F=0Ax^2 + Cy^2 + Dx + Ey + F = 0

Where AA and CC have the same sign but are not equal (otherwise, it would be a circle). Our given equation, x2+4y2−10x−40y+121=0x^2 + 4y^2 - 10x - 40y + 121 = 0, is in this general form. To find the center, we need to transform this general form into the standard form. The key technique to achieve this transformation is completing the square.

Step 1: Grouping Like Terms and Preparing to Complete the Square

Our initial equation is: x2+4y2−10x−40y+121=0x^2 + 4y^2 - 10x - 40y + 121 = 0. The first step involves grouping the xx terms together and the yy terms together. This sets the stage for completing the square for both variables separately.

Rearrange the equation as follows:

(x2−10x)+(4y2−40y)=−121(x^2 - 10x) + (4y^2 - 40y) = -121

Notice that we've moved the constant term (121) to the right side of the equation. This isolates the variable terms and prepares us for the next crucial step. Now, focus on the yy terms. Before we can complete the square for the yy terms, we need to factor out the coefficient of the y2y^2 term, which is 4 in this case. Factoring out this coefficient ensures that the coefficient of y2y^2 inside the parenthesis is 1, a necessary condition for completing the square.

(x2−10x)+4(y2−10y)=−121(x^2 - 10x) + 4(y^2 - 10y) = -121

We now have the xx and yy terms grouped and ready for the completion of the square process. Completing the square is a technique used to convert a quadratic expression into a perfect square trinomial, which can then be factored into a squared binomial. This is a fundamental algebraic manipulation that allows us to rewrite the equation in the standard form of an ellipse, revealing the center coordinates.

Step 2: Completing the Square for the x-terms

Completing the square is a vital technique for rewriting quadratic expressions. To complete the square for the xx terms (x2−10xx^2 - 10x), we need to add and subtract a specific value that will create a perfect square trinomial. This value is determined by taking half of the coefficient of the xx term (which is -10), squaring it, and adding it to the expression. Half of -10 is -5, and (-5) squared is 25. Therefore, we add and subtract 25 inside the parentheses.

(x2−10x+25−25)+4(y2−10y)=−121(x^2 - 10x + 25 - 25) + 4(y^2 - 10y) = -121

Now, the first three terms inside the parenthesis (x2−10x+25x^2 - 10x + 25) form a perfect square trinomial. We can rewrite this as a squared binomial:

((x−5)2−25)+4(y2−10y)=−121((x - 5)^2 - 25) + 4(y^2 - 10y) = -121

This step is crucial because it transforms the quadratic expression into a form that reveals the xx-coordinate of the center of the ellipse. We've successfully completed the square for the xx terms and are now ready to move on to the yy terms. Remember, the goal is to rewrite the entire equation in the standard form of an ellipse, and completing the square is a key step in achieving this.

Step 3: Completing the Square for the y-terms

Having successfully completed the square for the xx terms, we now turn our attention to the yy terms. We have the expression 4(y2−10y)4(y^2 - 10y). Similar to the xx terms, we need to add and subtract a value inside the parenthesis to create a perfect square trinomial. The coefficient of the yy term inside the parenthesis is -10. Half of -10 is -5, and (-5) squared is 25. So, we add and subtract 25 inside the parenthesis. It's crucial to remember that this 25 is inside the parenthesis which is being multiplied by 4, so we are effectively adding and subtracting 4∗254 * 25 to the left side of the equation.

((x−5)2−25)+4(y2−10y+25−25)=−121((x - 5)^2 - 25) + 4(y^2 - 10y + 25 - 25) = -121

Now, the first three terms inside the parenthesis involving yy (y2−10y+25y^2 - 10y + 25) form a perfect square trinomial. We can rewrite this as a squared binomial:

((x−5)2−25)+4((y−5)2−25)=−121((x - 5)^2 - 25) + 4((y - 5)^2 - 25) = -121

This step mirrors the process we used for the xx terms and is equally important. By completing the square for the yy terms, we are revealing the yy-coordinate of the center of the ellipse. We are one step closer to having the equation in standard form.

Step 4: Simplifying and Rearranging the Equation

With both the xx and yy terms in completed square form, we now simplify and rearrange the equation to get it closer to the standard form of an ellipse equation. We start by distributing the 4 in the second term and moving the constants to the right side of the equation.

(x−5)2−25+4(y−5)2−100=−121(x - 5)^2 - 25 + 4(y - 5)^2 - 100 = -121

Now, let's combine the constant terms on the left side:

(x−5)2+4(y−5)2=−121+25+100(x - 5)^2 + 4(y - 5)^2 = -121 + 25 + 100

This simplifies to:

(x−5)2+4(y−5)2=4(x - 5)^2 + 4(y - 5)^2 = 4

To get the equation into standard form, we need the right side to be equal to 1. So, we divide both sides of the equation by 4:

(x−5)24+(y−5)21=1\frac{(x - 5)^2}{4} + \frac{(y - 5)^2}{1} = 1

This is the standard form of the ellipse equation. By manipulating the equation through completing the square and simplifying, we have successfully transformed the general form into the standard form. This is a significant achievement, as the standard form directly reveals the key parameters of the ellipse, including its center.

Step 5: Identifying the Center of the Ellipse

Now that we have the equation in the standard form:

(x−5)24+(y−5)21=1\frac{(x - 5)^2}{4} + \frac{(y - 5)^2}{1} = 1

We can directly identify the center of the ellipse. Comparing this equation to the standard form (x−h)2a2+(y−k)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, we can see that:

  • h=5h = 5
  • k=5k = 5

Therefore, the center of the ellipse is the point (h,k)=(5,5)(h, k) = (5, 5). The center of the ellipse is a crucial characteristic, as it represents the central point around which the ellipse is symmetric. Knowing the center allows us to visualize the ellipse's position in the coordinate plane and understand its relationship to other geometric figures.

Conclusion: Center of the Ellipse

In conclusion, by employing the method of completing the square, we have successfully transformed the general equation of the ellipse, x2+4y2−10x−40y+121=0x^2 + 4y^2 - 10x - 40y + 121 = 0, into its standard form, (x−5)24+(y−5)21=1\frac{(x - 5)^2}{4} + \frac{(y - 5)^2}{1} = 1. From this standard form, we have definitively identified the center of the ellipse as (5, 5). This step-by-step guide provides a clear and concise method for finding the center of any ellipse given its equation. Understanding the process of completing the square and recognizing the standard form of an ellipse equation are essential skills in analytic geometry.

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