How do you graph the ellipse? (x - 6)²/36 + (y + 3)²/100 = 1

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How do you graph the ellipse x 636 y 3100 1 Drag choices into the boxes to correctly complete the statements class=

Respuesta :

Answer:

center: [tex](6, -3)[/tex]

distance from center:

  • major axis - 10 units
  • minor axis - 6 units

connect: [tex](16, -3), (6, -9), (-4, -3), (6, 3)[/tex]

Step-by-step explanation:

Ellipse Equation:

An ellipse has two forms, although they're essentially the same:

[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]

and

[tex]\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1[/tex]

Where a > b, and if the "a^2" is under the [tex](x-h)^2[/tex] then that means the ellipse has a horizontal major axis, but in general the value below this numerator relates to the length of the horizontal axis. Now if "a^2" is under the [tex](y-k)^2[/tex] then that means the ellipse has a vertical major axis, but in general the value below this numerator relates to the length of the vertical axis.

The length of the major axis is defined as 2a and the minor axis as 2b, but the distance from the center to the end points will be half of this, so either "a" or "b" depending on which axis the end point is on.

The other thing to note is that (h, k) is the center of the ellipse.

Analyzing the Equation:

We're given the equation: [tex]\frac{(x-6)^2}{36}+\frac{(y+3)^2}{100}=1[/tex], in this case the bigger value is under the (y+3)^2, so we know that we will have a horizontal major axis. Since the denominator's represent the square of "a" and "b" we first have to take the square root of them. So the following is true:

[tex]a=\sqrt{100}=10\\b=\sqrt{36}=6[/tex]

The "a" gives us the distance from the center to the end point on the major axis while the "b" gives us the distance from the center to the end point on the minor axis, so the distance from the center to the end point on the minor axis is 6 and for the major axis it's 10.

We also know the center is at (6, -3) by looking at the signs and what is being added/subtracted from the x and y values in the equation. We can use this to calculate the end points since the end points on the horizontal major axis can be calculated as: [tex](6\pm10, -3)[/tex] and the end points on the vertical major axis can be calculated as: [tex](6, -3\pm 6)[/tex], which gives us four points: [tex](16, -3), (6, -9), (-4, -3), (6, 3)[/tex] which we can connect to draw a graph.

Answer:

[tex]\textsf{The center of the ellipse is $\boxed{(6,-3)}$}\;.[/tex]

[tex]\textsf{The endpoints of the major axis are $\boxed{10}$ units from the center}\;.[/tex]

[tex]\textsf{The endpoints of the minor axis are $\boxed{6}$ units from the center}\;.[/tex]

[tex]\textsf{To graph the ellipse, connect $\boxed{(12,-3),(6,-13),(0,-3), \;\textsf{and}\; (6,7)}$ with a smooth curve}\;.[/tex]

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{7.2 cm}\underline{General equation of an ellipse}\\\\$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center\\ \phantom{ww}$\bullet$ $a$ and $b$ are the radii.\\ \phantom{ww}$\bullet$ $(h\pm a,k)$ and $(h,k\pm b)$ are the vertices.\\ \end{minipage}}[/tex]

Given equation:

[tex]\dfrac{(x-6)^2}{36}+\dfrac{(y+3)^2}{100}=1[/tex]

As b > a, the given ellipse is vertical.

The major axis is the longest diameter and the minor axis is the shortest diameter, therefore:

  • Major axis = 2b
  • Minor axis = 2a
  • Vertices = (h, k±b)
  • Co-vertices = (h±a, k)

Determine the values of h and k:

[tex](x-h)=(x-6) \implies h=6[/tex]

[tex](y-k)=(y+3) \implies k=-3[/tex]

Therefore, the center of the ellipse is:

  • [tex](h,k)=(6, -3)[/tex]

Calculate the values of a and b:

[tex]a^2=36 \implies a=6[/tex]

[tex]b^2=100 \implies b=10[/tex]

As the major axis is 2b, then the major radius is b.

Therefore, the endpoints of the major axis are "b" units from the center, so they are 10 units from the center.

As the minor axis is 2a, then the minor radius is a.

Therefore, the endpoints of the minor axis are "a" units from the center, so they are 6 units from the center.

To graph the ellipse, connect the vertices and co-vertices with a smooth curve.

  • Vertices = (h, k±b) = (6, -3±10) = (6, -13) and (6, 7)
  • Co-vertices = (h±a, k) = (6±6, -3) = (0, -3) and (12, -3)
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