<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>17.1.1.1 The Ellipse, Parabola and Hyperbola</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>The ellipse, parabola and hyperbola</h3>

<p>Mathematicians, engineers and scientists encounter numerous functions in their work: polynomials, trigonometric and hyperbolic functions amongst them. However, throughout the history of science one group of functions, the conics, arise time and time again
    not only in the development of mathematical theory but also in practical applications. The conics were first studied by the Greek mathematician Apollonius more than 200 years BC.</p>
<p>Essentially, the conics form that class of curves which are obtained when a double cone is intersected by a plane. There are three main types: the <strong>ellipse</strong>, the <strong>parabola</strong> and the <strong>hyperbola</strong>. From the ellipse
    we obtain the <strong>circle</strong> as a special case, and from the hyperbola we obtain the rectangular <strong>hyperbola</strong> as a special case. These curves are illustrated in the following figures.</p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/17-1%20Figer1a%20Circle%20and%20Ellipse.svg" alt="A figure showing how a circle and an ellipse are obtained when a double cone is intersected by a plane." width="616" height="419" class="img-responsive atto_image_button_text-bottom">
    <figcaption><strong>Figure:</strong> Circle and ellipse </figcaption>
</figure>
	
<hr>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/17-1%20Figure%201b%20Parabola.svg" alt="A figure showing how the parabola may be obtained from when a double cone is intersected by a plane." width="567" height="236" class="img-responsive atto_image_button_text-bottom">
</figure>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/17-1%20Figure%201c%20Hyperbola.svg" alt="Figure showing how an hyperbola is obtained when a double code is intersected by a plane." width="599" height="248" class="img-responsive atto_image_button_text-bottom">
    <figcaption> <strong>Figure:</strong>Parabola and hyperbola</figcaption>
</figure><br>


<h4>The ellipse</h4>
<p>We are all aware that the paths followed by the planets around the sun are elliptical. However, more generally the ellipse occurs in many areas of engineering. The standard form of an ellipse is shown below.</p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/17-1%20Figer2%20Ellipse%20standard%20form.svg" alt="The standard form of an ellipse" width="567" height="196" class="img-responsive atto_image_button_text-bottom">
</figure>
If \(a\gt b\) then the \(x\)-axis is called the <strong>major-axis</strong> and the \(y\)-axis is called the <strong>minor-axis</strong>. On the other hand if \(b\gt a\) then the \(y\)-axis is called the major-axis and the \(x\)-axis is then the minor-axis.
Two points inside the ellipse are of importance; these are the <strong>foci</strong>. If \(a\gt b\) these are located at coordinate positions \(\pm a e\) (or at \(\pm be\) if \(b\gt a\)) on the major-axis, with \(e\), called the <strong>eccentricity</strong>,
given by<br>\[e^2=1-\dfrac{b^2}{a^2}\qquad (b\lt a) \qquad\text{or by}\qquad e^2=1-\dfrac{a^2}{b^2}\qquad (a\lt b).\]<br>
<p>The foci of an ellipse have the property that if light rays are emitted from one focus then on reflection at the elliptic curve they pass through at the other focus.</p>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p>The standard Cartesian equation of the {\bf ellipse} with its centre at the origin is</p>
    <p>\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.\]</p>
</div>
This ellipse has intercepts on the \(x\)-axis at \(x=\pm a\) and on the \(y\)-axis at \(\pm b\). The curve is also symmetrical about both axes. The curve reduces to a circle in the special case in which \(a=b\).

<hr>
<h4 class="HELM_example">Example</h4>
<ol class="HELM_parts_inline">
    <li>Sketch the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{9} = 1\)</li>
    <li>Find the eccentricity \(e\).</li>
    <li>Locate the positions of the foci.</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>
        We can calculate the values of \(y\) as \(x\) changes from \(0\) to \(2\):
        <table style="text-align: center; border: 1px solid black; border-collapse: collapse;">
            <tbody>
                <tr style="border: 1px solid black; border-collapse: collapse;">
                    <td>&nbsp; \(x\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(0\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(0.30\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(0.60\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(0.90\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(1.20\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(1.50\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(1.80\)&nbsp;&nbsp;</td>
                    <td>&nbsp; \(2\)&nbsp;&nbsp;</td>
                </tr>
                <tr>
                    <td>\(y\)</td>
                    <td>\(3\)</td>
                    <td>\(2.97\)</td>
                    <td>\(2.86\)</td>
                    <td>\(2.68\)</td>
                    <td>\(2.40\)</td>
                    <td>\(1.98\)</td>
                    <td>\(1.31\)</td>
                    <td>\(0\)</td>
                </tr>
            </tbody>
        </table>
        From this table of values, and using the symmetry of the curve, a sketch can be drawn (see the figure below). Here \(b=3\) and \(a=2\) so the \(y\)-axis is the major axis and the \(x\)-axis is the minor axis.
    </li>
    <li>\(e^2 = 1-a^2/b^2 = 1-4/9 = 5/9\), \(\therefore e = \sqrt{5}/3\).</li>
    <li>Since \(b\gt a\) and \(be = \sqrt{5}\), the foci are located at \(\pm \sqrt{5}\) on the \(y\)-axis.
        <figure class="HELM_figure">
            <img src="@@PLUGINFILE@@/17-1%20figure%203%20example%201.svg" alt="The sketch associated to Example 1" width="367" height="207" class="img-responsive atto_image_button_text-bottom"></figure>
    </li>
</ol>
The Key Point gives the equation of the ellipse with its centre at the origin.  If the centre of the ellipse has coordinates \((\alpha, \beta)\) and still has its axes parallel to the \(x\)- and \(y\)-axes the standard equation becomes
\[\dfrac{(x-\alpha)^2}{a^2}+\dfrac{(y-\beta)^2}{b^2}=1.\]]]></text>
<file name="17-1 Figer1a Circle and Ellipse.svg" path="/" encoding="base64">[encoded data, 139884 characters]</file>
<file name="17-1 Figer2 Ellipse standard form.svg" path="/" encoding="base64">[encoded data, 53040 characters]</file>
<file name="17-1 Figure 1b Parabola.svg" path="/" encoding="base64">[encoded data, 66656 characters]</file>
<file name="17-1 Figure 1c Hyperbola.svg" path="/" encoding="base64">[encoded data, 148464 characters]</file>
<file name="17-1 figure 3 example 1.svg" path="/" encoding="base64">[encoded data, 30724 characters]</file>
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