Identifying Ellipses With Eccentricity Below 0.5

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In the realm of conic sections, the ellipse holds a prominent position, characterized by its unique shape and mathematical properties. Among these properties, eccentricity stands out as a crucial parameter that defines the elongation of an ellipse. An ellipse's eccentricity, denoted by e, is a non-negative number that determines how much the ellipse deviates from a perfect circle. The eccentricity e ranges from 0 to 1, where e = 0 represents a circle and e approaching 1 indicates a highly elongated ellipse. In this comprehensive guide, we will embark on a journey to identify ellipses represented by equations whose eccentricities are less than 0.5. This exploration will involve a step-by-step process of transforming the given equations into standard form, calculating the eccentricity, and ultimately determining whether the ellipse meets the specified criterion.

The provided equations are:

49x2βˆ’98x+64y2+256yβˆ’2,831=081x2βˆ’648x+100y2+200yβˆ’6,704=06x2βˆ’12x+54y2+108yβˆ’426=049x2+196x+36y2+216yβˆ’1,244=0\begin{array}{ll} 49 x^2-98 x+64 y^2+256 y-2,831=0 & 81 x^2-648 x+100 y^2+200 y-6,704=0 \\ 6 x^2-12 x+54 y^2+108 y-426=0 & 49 x^2+196 x+36 y^2+216 y-1,244=0 \end{array}

Our mission is to meticulously analyze each equation, unravel its underlying elliptical form, and compute its eccentricity. By comparing the calculated eccentricities with the threshold of 0.5, we will precisely identify the ellipses that satisfy the given condition. This process will not only enhance our understanding of ellipses but also equip us with the skills to analyze and classify conic sections based on their equations.

Step 1: Transforming Equations into Standard Form

The first crucial step in our endeavor is to transform the given equations into the standard form of an ellipse. The standard form provides a clear representation of the ellipse's key parameters, such as the center, major axis, and minor axis. This transformation involves a meticulous process of completing the square for both the x and y terms. Let's delve into the details of this process.

To effectively complete the square, we will systematically group the x terms and the y terms together, ensuring that the coefficients of the squared terms are equal to 1. This preparation paves the way for the application of the completing the square technique. By adding and subtracting appropriate constants, we will manipulate the equations to create perfect square trinomials in both x and y. These perfect square trinomials can then be elegantly factored into squared binomials, bringing us closer to the standard form.

The standard form of an ellipse equation is:

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

where:

  • (h, k) represents the center of the ellipse
  • a represents the semi-major axis (half the length of the major axis)
  • b represents the semi-minor axis (half the length of the minor axis)

By meticulously transforming the given equations into this standard form, we will unlock the essential parameters needed to calculate the eccentricity of each ellipse. This step is paramount in our quest to identify ellipses with eccentricities less than 0.5.

Let's apply this transformation to each of the given equations:

Equation 1: $49 x^2-98 x+64 y^2+256 y-2,831=0$

  1. Group x and y terms:

    (49x2βˆ’98x)+(64y2+256y)=2,831(49 x^2 - 98 x) + (64 y^2 + 256 y) = 2,831

  2. Factor out coefficients:

    49(x2βˆ’2x)+64(y2+4y)=2,83149(x^2 - 2x) + 64(y^2 + 4y) = 2,831

  3. Complete the square:
    • For x: (x2βˆ’2x+1)=(xβˆ’1)2(x^2 - 2x + 1) = (x - 1)^2, add 49βˆ—1=4949 * 1 = 49 to both sides.
    • For y: (y2+4y+4)=(y+2)2(y^2 + 4y + 4) = (y + 2)^2, add 64βˆ—4=25664 * 4 = 256 to both sides.

    49(x2βˆ’2x+1)+64(y2+4y+4)=2,831+49+25649(x^2 - 2x + 1) + 64(y^2 + 4y + 4) = 2,831 + 49 + 256

  4. Factor and simplify:

    49(xβˆ’1)2+64(y+2)2=3,13649(x - 1)^2 + 64(y + 2)^2 = 3,136

  5. Divide by 3136 to get the standard form:

    (xβˆ’1)264+(y+2)249=1\frac{(x - 1)^2}{64} + \frac{(y + 2)^2}{49} = 1

Equation 2: $81 x^2-648 x+100 y^2+200 y-6,704=0$

  1. Group x and y terms:

    (81x2βˆ’648x)+(100y2+200y)=6,704(81 x^2 - 648 x) + (100 y^2 + 200 y) = 6,704

  2. Factor out coefficients:

    81(x2βˆ’8x)+100(y2+2y)=6,70481(x^2 - 8x) + 100(y^2 + 2y) = 6,704

  3. Complete the square:
    • For x: (x2βˆ’8x+16)=(xβˆ’4)2(x^2 - 8x + 16) = (x - 4)^2, add 81βˆ—16=129681 * 16 = 1296 to both sides.
    • For y: (y2+2y+1)=(y+1)2(y^2 + 2y + 1) = (y + 1)^2, add 100βˆ—1=100100 * 1 = 100 to both sides.

    81(x2βˆ’8x+16)+100(y2+2y+1)=6,704+1,296+10081(x^2 - 8x + 16) + 100(y^2 + 2y + 1) = 6,704 + 1,296 + 100

  4. Factor and simplify:

    81(xβˆ’4)2+100(y+1)2=8,10081(x - 4)^2 + 100(y + 1)^2 = 8,100

  5. Divide by 8100 to get the standard form:

    (xβˆ’4)2100+(y+1)281=1\frac{(x - 4)^2}{100} + \frac{(y + 1)^2}{81} = 1

Equation 3: $6 x^2-12 x+54 y^2+108 y-426=0$

  1. Group x and y terms:

    (6x2βˆ’12x)+(54y2+108y)=426(6 x^2 - 12 x) + (54 y^2 + 108 y) = 426

  2. Factor out coefficients:

    6(x2βˆ’2x)+54(y2+2y)=4266(x^2 - 2x) + 54(y^2 + 2y) = 426

  3. Complete the square:
    • For x: (x2βˆ’2x+1)=(xβˆ’1)2(x^2 - 2x + 1) = (x - 1)^2, add 6βˆ—1=66 * 1 = 6 to both sides.
    • For y: (y2+2y+1)=(y+1)2(y^2 + 2y + 1) = (y + 1)^2, add 54βˆ—1=5454 * 1 = 54 to both sides.

    6(x2βˆ’2x+1)+54(y2+2y+1)=426+6+546(x^2 - 2x + 1) + 54(y^2 + 2y + 1) = 426 + 6 + 54

  4. Factor and simplify:

    6(xβˆ’1)2+54(y+1)2=4866(x - 1)^2 + 54(y + 1)^2 = 486

  5. Divide by 486 to get the standard form:

    (xβˆ’1)281+(y+1)29=1\frac{(x - 1)^2}{81} + \frac{(y + 1)^2}{9} = 1

Equation 4: $49 x^2+196 x+36 y^2+216 y-1,244=0$

  1. Group x and y terms:

    (49x2+196x)+(36y2+216y)=1,244(49 x^2 + 196 x) + (36 y^2 + 216 y) = 1,244

  2. Factor out coefficients:

    49(x2+4x)+36(y2+6y)=1,24449(x^2 + 4x) + 36(y^2 + 6y) = 1,244

  3. Complete the square:
    • For x: (x2+4x+4)=(x+2)2(x^2 + 4x + 4) = (x + 2)^2, add 49βˆ—4=19649 * 4 = 196 to both sides.
    • For y: (y2+6y+9)=(y+3)2(y^2 + 6y + 9) = (y + 3)^2, add 36βˆ—9=32436 * 9 = 324 to both sides.

    49(x2+4x+4)+36(y2+6y+9)=1,244+196+32449(x^2 + 4x + 4) + 36(y^2 + 6y + 9) = 1,244 + 196 + 324

  4. Factor and simplify:

    49(x+2)2+36(y+3)2=1,76449(x + 2)^2 + 36(y + 3)^2 = 1,764

  5. Divide by 1764 to get the standard form:

    (x+2)236+(y+3)249=1\frac{(x + 2)^2}{36} + \frac{(y + 3)^2}{49} = 1

Step 2: Calculating Eccentricity

With the equations now in standard form, the next pivotal step is to calculate the eccentricity (e) of each ellipse. Eccentricity, as we discussed earlier, serves as a crucial measure of an ellipse's elongation. The formula for calculating eccentricity is:

e=1βˆ’b2a2e = \sqrt{1 - \frac{b^2}{a^2}}

where:

  • a represents the semi-major axis (the larger of the two axes)
  • b represents the semi-minor axis (the smaller of the two axes)

This formula elegantly captures the relationship between the axes and the eccentricity, providing a quantitative measure of how