/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Find an equation for the ellipse... [FREE SOLUTION] | 91Ó°ÊÓ

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Find an equation for the ellipse with foci at \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\) that passes through \((0,3)\)

Short Answer

Expert verified
The equation is \(\frac{x^2}{12} + \frac{y^2}{9} = 1\).

Step by step solution

01

Understanding the Ellipse Parameters

An ellipse centered at the origin is generally defined by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis, \(b\) is the semi-minor axis, and for the ellipse with horizontal major axis, \(c\) is the distance from the center to each focus (\(c \leq a\)). Here, the foci are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\), so \(c = \sqrt{3}\).
02

Relate the Foci to Semi-Major Axis

We know the relationship \(c^2 = a^2 - b^2\). Given \(c = \sqrt{3}\), we have \(c^2 = 3 = a^2 - b^2\).
03

Use the Point to Find the Relationship

The ellipse passes through the point \((0,3)\), so substitute \((0,3)\) into \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\): \(\frac{0^2}{a^2} + \frac{3^2}{b^2} = 1\). Hence, \(\frac{9}{b^2} = 1\), which implies \(b^2 = 9\).
04

Solve for \(a^2\)

Using \(c^2 = a^2 - b^2\), where \(b^2 = 9\), substitute to get \(3 = a^2 - 9\). Thus, solving for \(a^2\) gives \(a^2 = 12\).
05

Formulate the Ellipse Equation

Substitute \(a^2 = 12\) and \(b^2 = 9\) back into the ellipse equation: \(\frac{x^2}{12} + \frac{y^2}{9} = 1\). Hence, the equation for the ellipse is \(\frac{x^2}{12} + \frac{y^2}{9} = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Semi-Major Axis
The semi-major axis of an ellipse is the longest radius from the center to the edge of the ellipse. In simpler terms, it's half the length of the longest line that can be drawn through the center of the ellipse. This concept is crucial in understanding the shape and size of an ellipse. When reading the standard ellipse equation, \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the term \(a\) represents the semi-major axis. If \(a > b\), the ellipse stretches more along the x-axis. Conversely, if \(a < b\), it stretches more along the y-axis.
In our given problem, we found \(a^2 = 12\), which means \(a = \sqrt{12} = 2\sqrt{3}\). This tells us how far the ellipse extends horizontally across the x-axis. Understanding the semi-major axis helps in visualizing the overall shape of the ellipse, emphasizing its stretched dimension.
Semi-Minor Axis
The semi-minor axis is the shortest radius from the center of the ellipse to its edge. Essentially, it's half the length of the shortest line that runs through the center of the ellipse. In the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the term \(b\) stands for the semi-minor axis. It indicates the extent of the ellipse along the y-axis or vertical direction if \(a > b\).
In the exercise, the value of \(b^2\) was determined to be \(9\), giving \(b = \sqrt{9} = 3\). This means that the total vertical height (top to bottom) of the ellipse is twice the semi-minor axis, 6 in this instance. Recognizing \(b\) into the context of the ellipse’s structure allows one to appreciate how the ellipse 'squashes' along the shorter axis, shaping its oval appearance.
Foci of Ellipse
The foci (plural for focus) of an ellipse are two special points located along the major axis. The sum of the distances from each focus to any point on the ellipse remains constant. This property is what fundamentally defines an ellipse’s shape.
In our exercise, these foci were at \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\), indicating that the major axis is horizontal. The distance from the center to either focus, denoted as \(c\), is vital. Here, \(c = \sqrt{3}\).
To further grasp this concept, remember that for any ellipse, \(c^2 = a^2 - b^2\). This equation ties the foci with the semi-major and semi-minor axes, offering deeper insight into the ellipse's geometry. In visual terms, if you picture an ellipse as a stretched circle, the foci remain the pivotal points that dictate the extent of that stretch, anchoring the ellipse's structure.
Coordinate Geometry
Coordinate geometry, often termed as analytic geometry, involves studying geometric figures using a coordinate system. For ellipses, coordinate geometry provides the tools to represent curves in a plane, with our primary focus on the Cartesian coordinate system.
Through this system, an ellipse can be described via equations that pinpoint its size and shape precisely. The general equation of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) allows us to derive important characteristics like the axes and foci from coordinates accurately.
In solving our ellipse problem, coordinate geometry helps us use set coordinates to define the ellipse's distinctive paths and dimensions. By substituting and solving these coordinates, we navigate smoothly through determining elements like the semi-major and semi-minor axes, reinforcing how algebra and geometry coalesce to unravel such spatial relations.

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Most popular questions from this chapter

CHALLENGE For Exercises \(44-48,\) find all values of \(k\) for which the system of equations has the given number of solutions. If no values of \(k\) meet the condition, write none. $$ x^{2}+y^{2}=k^{2} \quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 $$ no solutions

For Exercises \(34-37,\) use the following information. A hyperbola with asymptotes that are not perpendicular is called a nonrectangular hyperbola. Most of the hyperbolas you have studied so far are nonrectangular. A rectangular hyperbola is a hyperbola with perpendicular asymptotes. For example, the graph of \(x^{2}-y^{2}=1\) is a rectangular hyperbola. The graphs of equations of the form \(x y=c,\) where \(c\) is a constant, are rectangular hyperbolas with the coordinate axes as their asymptotes. Find the coordinates of the vertices of the graph of \(x y=2\)

Write an equation for the ellipse that satisfies each set of conditions. foci at \((5,4)\) and \((-3,4),\) major axis 10 units long

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ \frac{x^{2}}{9}-\frac{y^{2}}{25}=1 $$

A curved mirror is placed in a store for a wide-angle view of the room. The equation \(\frac{x^{2}}{1}-\frac{y^{2}}{3}=1\) models the curvature of the mirror. A small security camera is placed 3 feet from the vertex of the mirror so that a diameter of 2 feet of the mirror is visible. If the back of the room lies on \(x=-18\) , what width of the back of the room is visible to the camera?

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