Problem 30
Question
Write an equation of an ellipse for the given foci and co-vertices. foci \((0, \pm 4),\) co-vertices \(( \pm 2,0)\)
Step-by-Step Solution
Verified Answer
The equation of the ellipse is \(\frac{x^2}{4} + \frac{y^2}{20} = 1\).
1Step 1: Identify given information
The given information is that the foci are at (0,±4) and the co-vertices are at (±2,0). This means c=4 and b=2.
2Step 2: Calculate the semi-major axis
Use the equation \(a^2 = b^2 + c^2\) to calculate the semi-major axis. In this case, \(a^2 = 2^2 + 4^2 = 20\), hence a is the square root of 20 or in simplified form \(2\sqrt{5}\).
3Step 3: Construct the equation of the ellipse
The equation of the ellipse when foci are on y-axis and the centre is at the origin is given by \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\). Substituting b=2 and a=\(2\sqrt{5}\) we get the equation \(\frac{x^2}{4} + \frac{y^2}{20} = 1\).
Key Concepts
Foci of an EllipseCo-vertices and Their RoleUnderstanding the Semi-Major AxisExploring Ellipse Geometry
Foci of an Ellipse
The foci of an ellipse are key in defining its shape and position. They are two distinct points located along the major axis. For the ellipse given in the exercise, the foci are at the coordinates
- (0, 4) and (0, -4)
- (0, 0)
- c = 4
Co-vertices and Their Role
Co-vertices are another important aspect of ellipse geometry. These points lie on the ellipse's minor axis, perpendicular to the major axis. For the exercise at hand, the co-vertices are located at
- (±2, 0)
- b = 2
Understanding the Semi-Major Axis
The semi-major axis of an ellipse is the longest radius extending from the center to the edge along the major axis. In this exercise, the semi-major axis is calculated using the formula:
- \[ a^2 = b^2 + c^2 \]
- b = 2
- c = 4
- \[ a^2 = 2^2 + 4^2 = 20 \]
Exploring Ellipse Geometry
Ellipse geometry is a fascinating area of mathematics, heavily reliant on understanding its axes, foci, and vertices. The standard equation of an ellipse with a vertical major axis centered at the origin is
- \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]
- a = 2\sqrt{5}
- b = 2
- \[ \frac{x^2}{4} + \frac{y^2}{20} = 1 \]
Other exercises in this chapter
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