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Chapter 11: Problem 4

Sketch the three basic conic sections in standard position with vertices and foci on the \(x\) -axis.

Short Answer

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Question: Sketch the three basic conic sections (ellipse, parabola, and hyperbola) in their standard positions with vertices and foci on the x-axis. Provide the vertices and foci for each conic section. Answer: Ellipse: The vertices are at (卤3, 0), and the foci are at (卤鈭�5, 0). Parabola: The vertex is at (0, 0), and the focus point is at (0, 1/4). Hyperbola: The vertices are at (卤3, 0), and the foci are at (卤鈭�13, 0).

Step by step solution

01

Standard form of an ellipse

An ellipse is represented by this standard form of equation: \[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\] In this case, since the vertex and foci should lie on the x-axis, the horizontal axis is larger than the vertical axis, so \(a > b\). Also, the center of the ellipse will be at the origin, so \(h = 0\) and \(k = 0\).
02

Ellipse vertices and foci

With the center of the ellipse at the origin, the vertices can be determined as (\(\pm{a}, 0\)). In this case, let us assume that \(a = 3\) and \(b= 2\). Hence, the vertices are at (\(\pm 3, 0\)). The equation for finding the foci distance from the center is: \(c = \sqrt{a^2 - b^2}\). For our ellipse, \(c = \sqrt{3^2 - 2^2} = \sqrt{5}\). Thus, the points for the foci are at (\(\pm \sqrt{5}, 0\)). #Step 2: Parabola#
03

Standard form of a parabola

A parabola is represented by this standard form of equation: \[y = a(x-h)^2+k\] In this case, the vertex will be located on the x-axis so \(k = 0\). Let's keep the vertex at the origin, so \(h = 0\) as well.
04

Parabola vertex and focus point

Given the origin as the vertex (\(0, 0\)), and using the standard form of the equation, we simplify the equation to: \(y = ax^2\). Now, in order to find the focus point we can use the formula \(a = \frac{1}{4f}\), where \(f\) is the distance between the vertex and the focus point. For our case, let's assume that \(a = 1\), so the distance to the focus point would be \(4f = 1\), resulting in \(f = \frac{1}{4}\). Consequently, the focus point will be (\(0, \frac{1}{4}\)). #Step 3: Hyperbola#
05

Standard form of a hyperbola

A hyperbola is represented by this standard form of equation: \[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\] In this case, the center of the hyperbola will be at the origin, so \(h = 0\) and \(k = 0\).
06

Hyperbola vertices and foci

Given the center of the hyperbola at the origin, the vertices can be determined as (\(\pm{a}, 0\)). In this case, let us assume that \(a = 3\) and \(b= 2\). Hence, the vertices are at (\(\pm 3, 0\)). The equation for finding the foci distance from the center is: \(c = \sqrt{a^2 + b^2}\). For our hyperbola, \(c = \sqrt{3^2+2^2} = \sqrt{13}\). Thus, the points for the foci are at (\(\pm \sqrt{13}, 0\)).

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