Skip to content
This repository has been archived by the owner on May 11, 2021. It is now read-only.

Fixes for the hyperbola intuition exercise #161553

Open
wants to merge 5 commits into
base: master
Choose a base branch
from
Open

Fixes for the hyperbola intuition exercise #161553

Changes from all commits
Commits
Show all changes
5 commits
Select commit Hold shift + click to select a range
f92a15b
Make the final equation describe a hyperbola with the correct center.
wrwrwr Nov 25, 2015
be40a65
Swap semi-major and semi-minor axes to match the graph and references.
wrwrwr Nov 25, 2015
a50be01
Rename variables to make the east-west hyperbola code more readable.
wrwrwr Nov 25, 2015
f27ec26
Change hyperbola_intuition title to match its content.
wrwrwr Nov 25, 2015
ed27ed5
Fix the equation in the intro to the south-north hyperbola case.
wrwrwr Nov 25, 2015
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
58 changes: 29 additions & 29 deletions exercises/hyperbola_intuition.html
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2,7 +2,7 @@
<html data-require="math math-format graphie interactive parabola-intuition">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Hyperbola and ellipse intuition</title>
<title>Hyperbola intuition</title>
<script data-main="../local-only/main.js" src="../local-only/require.js"></script>
</head>
<body>
Expand All @@ -20,7 +20,7 @@

<div class="question">
<p>The equation of a north-south opening hyperbola is :<br/>
<code>\qquad \dfrac{(x - x_1)^2}{b^2} - \dfrac{(y - y_1)^2}{a^2} = 1</code>.
<code>\qquad \dfrac{(y - y_1)^2}{a^2} - \dfrac{(x - x_1)^2}{b^2} = 1</code>.
</p>
<p>
Find the foci of the hyperbola below by moving the orange points to their correct positions.
Expand Down Expand Up @@ -54,7 +54,7 @@
graph.H = H;
graph.K = K;
graph.C = C;

initAutoscaledGraph([[-20, 20], [-20, 20]], {});
addMouseLayer();

Expand Down Expand Up @@ -161,7 +161,7 @@
<p>The center of a hyperbola is at the midpoint of its two foci.</p>
<p><code>\left(\dfrac{<var>H</var> + <var>H</var>}{2}, \dfrac{<var>K+C</var> + <var>K-C</var>}{2}\right) = (<var>H</var>, <var>K</var>)</code></p>
</div>
<p>The absolute value of the distance from one focus minus the distance to the other focus is equal to <code>2a</code></p>
<p>The absolute value of the distance from one focus minus the distance to the other focus is equal to <code>2a</code>.</p>
<div>
<code>\qquad\begin{align*}
2a &amp;= <var>2 * A</var> \\
Expand Down Expand Up @@ -191,18 +191,18 @@
<var id="A, B, C">randFromArray([[3, 4, 5], [6, 8, 10], [9, 12, 15], [5, 12, 13]])</var>
<var id="A2">A * A</var>
<var id="B2">B * B</var>
<var id="MAX_Y">18 - C</var>
<var id="H">randRange(-5, 5)</var>
<var id="K">randRange(-MAX_Y, MAX_Y)</var>
<var id="MAX_X">18 - C</var>
<var id="H">randRange(-MAX_X, MAX_X)</var>
<var id="K">randRange(-5, 5)</var>
</div>

<div class="question">
<p>The equation of an east-west opening hyperbola is:<br/>
<code>\qquad \dfrac{(x - x_1)^2}{b^2} - \dfrac{(y - y_1)^2}{a^2} = 1</code>.
<code>\qquad \dfrac{(x - x_1)^2}{a^2} - \dfrac{(y - y_1)^2}{b^2} = 1</code>.
</p>
<p>
Find the foci of the hyperbola below by moving the orange points to their correct positions.
Then use that information to find the values of <code>y_1</code>, <code>x_1</code>, <code>a</code> and <code>b</code>.
Then use that information to find the values of <code>x_1</code>, <code>y_1</code>, <code>a</code> and <code>b</code>.
</p>
</div>

Expand Down Expand Up @@ -232,7 +232,7 @@
graph.H = H;
graph.K = K;
graph.C = C;

initAutoscaledGraph([[-20, 20], [-20, 20]], {});
addMouseLayer();

Expand All @@ -258,12 +258,12 @@
$("#problemarea span.focus2-y-label").html("&lt;code&gt;" + coordY + "&lt;/code&gt;").tex();
};

graph.func1 = addInteractiveFn(function(x) {
return K + sqrt(A2 / B2 * (x - H) * (x - H) + A2);
graph.func1 = addInteractiveFn(function(y) {
return H + sqrt(A2 / B2 * (y - K) * (y - K) + A2);
}, { swapAxes: true });

graph.func2 = addInteractiveFn(function(x) {
return K - sqrt(A2 / B2 * (x - H) * (x - H) + A2);
graph.func2 = addInteractiveFn(function(y) {
return H - sqrt(A2 / B2 * (y - K) * (y - K) + A2);
}, { swapAxes: true });

doHyperbolaInteraction(graph.func1, graph.focus1, graph.focus2);
Expand All @@ -290,9 +290,9 @@
return "";
}

return guess[0][1] === H &amp;&amp; guess[1][1] === H &amp;&amp;
(guess[0][0] === K - C &amp;&amp; guess[1][0] === K + C) ||
(guess[0][0] === K + C &amp;&amp; guess[1][0] === K - C)
return guess[0][1] === K &amp;&amp; guess[1][1] === K &amp;&amp;
(guess[0][0] === H - C &amp;&amp; guess[1][0] === H + C) ||
(guess[0][0] === H + C &amp;&amp; guess[1][0] === H - C)
</div>
<div class="show-guess">
graph.focus1.setCoord(guess[0]);
Expand All @@ -305,20 +305,20 @@
<table class="intuition-equation">
<tr>
<td><code>x_1 = </code></td>
<td><span class="sol short40" data-fallback="0"><var>K</var></span></td>
</tr>
<tr>
<td><code>y_1 = </code></td>
<td><span class="sol short40" data-fallback="0"><var>H</var></span></td>
</tr>
<tr>
<td><code>b = </code></td>
<td><span class="sol short40"><var>B</var></span></td>
<td><code>y_1 = </code></td>
<td><span class="sol short40" data-fallback="0"><var>K</var></span></td>
</tr>
<tr>
<td><code>a = </code></td>
<td><span class="sol short40"><var>A</var></span></td>
</tr>
<tr>
<td><code>b = </code></td>
<td><span class="sol short40"><var>B</var></span></td>
</tr>
</table>
</p>
</div>
Expand All @@ -329,17 +329,17 @@
There is only one way to arrange the two foci such that this is true.
<button onclick="javascript:
graph = KhanUtil.currentGraph.graph;
graph.focus1.moveTo(graph.K + graph.C, graph.H);
graph.focus2.moveTo(graph.K - graph.C, graph.H);
graph.focus1.moveTo(graph.H + graph.C, graph.K);
graph.focus2.moveTo(graph.H - graph.C, graph.K);
">Show me</button>
</p>
<p>One focus is <code>(<var>K - C</var>, <var>H</var>)</code> and the other is <code>(<var>K + C</var>, <var>H</var>)</code>.</p>
<p>One focus is <code>(<var>H - C</var>, <var>K</var>)</code> and the other is <code>(<var>H + C</var>, <var>K</var>)</code>.</p>
<p><code>x_1</code> and <code>y_1</code> are the coordinates of the center of the hyperbola.</p>
<div>
<p>The center of a hyperbola is at the midpoint of its two foci.</p>
<p><code>\left(\dfrac{<var>K - C</var> + <var>K + C</var>}{2}, \dfrac{<var>H</var> + <var>H</var>}{2}\right) = (<var>K</var>, <var>H</var>)</code></p>
<p><code>\left(\dfrac{<var>H - C</var> + <var>H + C</var>}{2}, \dfrac{<var>K</var> + <var>K</var>}{2}\right) = (<var>H</var>, <var>K</var>)</code></p>
</div>
<p>The absolute value of the distance from one focus minus the distance to the other focus is equal to <code>2a</code></p>
<p>The absolute value of the distance from one focus minus the distance to the other focus is equal to <code>2a</code>.</p>
<div>
<code>\qquad\begin{align*}
2a &amp;= <var>2 * A</var> \\
Expand All @@ -359,7 +359,7 @@
</div>
<p>
So the equation of the hyperbola is
<code>\dfrac{(x - <var>H</var>)^2}{<var>B</var>^2} + \dfrac{(y - <var>K</var>)^2}{<var>A</var>^2} = 1</code>.
<code>\dfrac{(x - <var>H</var>)^2}{<var>A</var>^2} - \dfrac{(y - <var>K</var>)^2}{<var>B</var>^2} = 1</code>.
</p>
</div>
</div>
Expand Down