Math Problem Statement

Find the standard form of an ellipse if the co vertices is (4, 8) and (10, 8), a focus at (3, 12) center with (7, 8)

Solution

We are given the following information for an ellipse:

  • Co-vertices: (4,8)(4, 8) and (10,8)(10, 8)
  • One focus: (3,12)(3, 12)
  • Center: (7,8)(7, 8)

Step 1: Determine the center of the ellipse

The center of the ellipse is directly given as (7,8)(7, 8).

Step 2: Determine the distance from the center to the co-vertices

The co-vertices are at (4,8)(4, 8) and (10,8)(10, 8). These lie along the minor axis of the ellipse. The minor axis is horizontal because both points have the same yy-coordinate, so the distance between the co-vertices is the length of the minor axis.

The distance between the center and a co-vertex is the semi-minor axis bb: b=1042=3b = \frac{10 - 4}{2} = 3 Thus, the length of the semi-minor axis is b=3b = 3.

Step 3: Determine the distance from the center to the focus

The focus lies at (3,12)(3, 12). The distance between the center (7,8)(7, 8) and the focus is the distance along the major axis, which we can calculate using the distance formula: c=(73)2+(812)2=(4)2+(4)2=16+16=32=42c = \sqrt{(7 - 3)^2 + (8 - 12)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} Thus, c=42c = 4\sqrt{2}.

Step 4: Use the relationship between aa, bb, and cc

In an ellipse, the relationship between the semi-major axis aa, semi-minor axis bb, and the focal distance cc is given by: c2=a2b2c^2 = a^2 - b^2 Substitute c=42c = 4\sqrt{2} and b=3b = 3 into this equation: (42)2=a232(4\sqrt{2})^2 = a^2 - 3^2 32=a2932 = a^2 - 9 a2=41a^2 = 41 Thus, a=41a = \sqrt{41}.

Step 5: Write the standard form of the equation

The standard form of the equation of an ellipse is: (xh)2a2+(yk)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 where (h,k)(h, k) is the center, aa is the length of the semi-major axis, and bb is the length of the semi-minor axis.

Substitute h=7h = 7, k=8k = 8, a2=41a^2 = 41, and b2=9b^2 = 9: (x7)241+(y8)29=1\frac{(x - 7)^2}{41} + \frac{(y - 8)^2}{9} = 1

Thus, the equation of the ellipse in standard form is: (x7)241+(y8)29=1\boxed{\frac{(x - 7)^2}{41} + \frac{(y - 8)^2}{9} = 1}

Would you like more details or have any other questions?

Here are 5 related questions to expand your understanding:

  1. How do you find the foci of an ellipse if the equation is given?
  2. What is the geometric interpretation of aa, bb, and cc in an ellipse?
  3. How do you identify the major and minor axes from an ellipse's equation?
  4. Can an ellipse have its major axis along the yy-axis instead of the xx-axis?
  5. What are the key differences between an ellipse and a hyperbola?

Tip: Always verify the center and orientation of an ellipse from its co-vertices and foci before writing the equation.

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Math Problem Analysis

Mathematical Concepts

Conic Sections
Ellipse
Distance Formula
Ellipse Equation

Formulas

Standard form of an ellipse: (x - h)^2/a^2 + (y - k)^2/b^2 = 1
Distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Ellipse relationship: c^2 = a^2 - b^2

Theorems

Ellipse Equation Theorem

Suitable Grade Level

Grades 10-12