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In this video, we’ll learn how to write the equation of a hyperbola and find the hyperbola’s vertices and foci using its equation and graph.
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We call it the conic sections or the curves we get when we intersect a cone with a plane.
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In the conic sections are the ellipse, of which the circle is a special case, the parabola, and the hyperbola.
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And the hyperbola, which is the curve we’re concerned with, consists of the two mirror-image curves resulting from a plane intersecting both halves of the cone.
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To give us an idea of what the equation of the hyperbola represents, it’s helpful to compare the components of an ellipse and a hyperbola.
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So let’s do that and then look more closely at the equation of a hyperbola and work through some examples.
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Remember, an ellipse is a plane curve defined by the equation 𝑥 squared over 𝑎 squared plus 𝑦 squared over 𝑏 squared is equal to one.
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And for any point on the curve, the sum of the distances from its two foci is a constant.
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And remember, a circle is a particular type of ellipse where the two focal points are equal.
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A noncircular ellipse has two lines of symmetry called the major and the minor axes.
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And the foci sit on the major axis which is the longer of the two.
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A hyperbola also has major and minor axes, and these are, again, lines of symmetry.
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And again, the foci lie on the major axis.
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But how an ellipse and a hyperbola differ is that in defining the locus of points, that’s the set of all points satisfying its equation, it’s not the sum of the distances from the foci that’s constant, but the difference of the distances from the point to the two foci.
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The hyperbola we have here is in what’s called standard position.
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And that’s where the center 𝑐 is at the origin, where 𝑥 is zero and 𝑦 is zero.
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Notice also that, in this case, the 𝑥-squared term is positive.
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And we’d say in this case that the hyperbola is horizontal or east-west opening; that is, the two branches are east and west facing.
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An east-west opening or a horizontal hyperbola has its major axis parallel to the 𝑥-axis.
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And as we’ve noted, that means the 𝑥-squared term is positive.
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The hyperbola shown here is a north south–opening hyperbola.
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And in this case, the 𝑦-squared term is positive.
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So the branches face north and the south, and it’s a vertical hyperbola.
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And in this case, the center is the origin, so it’s in standard position.
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In a north south–opening hyperbola, the major axis is parallel to the 𝑦-axis.
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Now, we’re going to concentrate on our east-west hyperbola, which has equation 𝑥 squared over 𝑎 squared minus 𝑦 squared over 𝑏 squared is equal to one, where 𝑎 and 𝑏 are greater than zero.
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The two points closest to the foci on the major axis are called the vertices.
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And their coordinates are negative 𝑎, zero and 𝑎, zero, so that the distance between the two vertices is two 𝑎.
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And the foci or foci are at the points 𝐹 one is equal to negative 𝑐, zero and 𝐹 two is equal to 𝑐, zero.
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And 𝑐 squared is equal to 𝑎 squared plus 𝑏 squared.
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Another important feature of the hyperbola is the asymptotes.
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These are two lines which pass through the center and which the branches get closer and closer to as 𝑥 gets larger and larger, although the branches never touch the asymptotes.
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These are the lines 𝑦 is equal to 𝑏 over 𝑎 𝑥 and 𝑦 is equal to negative 𝑏 over 𝑎 times 𝑥.
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So, in fact, these are the lines with slopes 𝑏 over 𝑎 and negative 𝑏 over 𝑎.
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And remember, this is an east west–opening hyperbola, which is centered at the origin.
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It’s worth noting that for a north south–opening hyperbola with equation 𝑦 squared over 𝑎 squared minus 𝑥 squared over 𝑏 squared is equal to one, we need to be careful in redefining our vertices, foci, and asymptotes.
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In this case, our vertices of 𝑉 one is equal to zero, negative 𝑎 and 𝑉 two is zero, 𝑎.
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The foci have coordinates zero, negative 𝑐 and zero, 𝑐.
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And the asymptotes are the lines 𝑦 is equal to plus or minus 𝑎 over 𝑏 𝑥.
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So now let’s look at how we might derive the equation for the east west–opening hyperbola in standard position.
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And remember that for a hyperbola, it’s the difference in distances from the foci to any point on the curve that’s constant.
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Now, 𝑉 two is a point on our curve with coordinates 𝑎, zero.
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So let’s take this as our point and look at the distance of this from the two foci.
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If we take 𝐷 one to be the distance from the vertex 𝑉 two to the first focal point 𝐹 one, that’s equal to 𝑎 plus 𝑐.
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And now if we take 𝐷 two to be the distance from 𝑉 two to the second focal point 𝐹 two, that’s equal to 𝑐 minus 𝑎.
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So now, if we take the difference of these two distances, that’s 𝑎 plus 𝑐 minus 𝑐 minus 𝑎.
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Our 𝑐’s cancel each other out, so we’re left with two 𝑎.
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And since the difference in distances from the foci of any point is constant, this must be our constant.
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So now for any point 𝑃: 𝑥, 𝑦 on the hyperbola, the difference between the distances from the two foci, 𝑑 one minus 𝑑 two, is equal to positive or negative two 𝑎.
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And we need the positive or negative because the sign of 𝑑 one minus 𝑑 two could be either, depending on which branch the point sits.
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From our graph, we see that we can use the Pythagorean theorem to find 𝑑 one and 𝑑 two.
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If the base of our triangle with hypotenuse 𝑑 one is 𝑐 plus 𝑥, then 𝑑 one is the square root of 𝑥 plus 𝑐 squared plus 𝑦 squared.
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And if the base of our triangle with hypotenuse 𝑑 two is 𝑐 minus 𝑥, then 𝑑 two is the square root of 𝑐 minus 𝑥 squared plus 𝑦 squared.
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And since 𝑐 minus 𝑥 squared is the same as 𝑥 minus 𝑐 squared, and so 𝑑 one minus 𝑑 two is the square root of 𝑥 plus 𝑐 squared plus 𝑦 squared minus the square root of 𝑥 minus 𝑐 squared plus 𝑦 squared.
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And that’s equal to positive or negative two 𝑎.
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Now, remember, we’re trying to derive this equation, and this is gonna take a little bit of algebra.
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So I’m going to make some space here and swap sides for our expressions.
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Now, if we add this second term to both sides, then on the left-hand side, this term cancels itself out.
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And now if we square both sides, we have 𝑥 plus 𝑐 squared plus 𝑦 squared on the left-hand side and our expanded bracket on the right-hand side.
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And if we multiply out these two brackets and collect like terms, we have 𝑥 times 𝑐 minus 𝑎 squared on the left-hand side and positive or negative 𝑎 times the square root of 𝑥 minus 𝑐 squared plus 𝑦 squared on the right-hand side.
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So now just making some space and rewriting that, if we square both sides again and expand our bracket inside the square root, multiplying out the bracket on the right-hand side, we can eliminate negative two 𝑎 squared 𝑥𝑐 on both sides by adding two 𝑎 squared 𝑥𝑐.
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And by taking 𝑎 raised to the fourth power to the right-hand side and 𝑎 squared 𝑥 squared and 𝑎 squared 𝑦 squared to the left-hand side, this reduces to 𝑐 squared minus 𝑎 squared times 𝑥 squared minus 𝑎 squared 𝑦 squared is equal to 𝑎 squared times 𝑐 squared minus 𝑎 squared.
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And now dividing everything by 𝑎 squared times 𝑐 squared minus 𝑎 squared, we can make quite a few cancellations.
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And we’re left with 𝑥 squared over 𝑎 squared minus 𝑦 squared over 𝑐 squared minus 𝑎 squared is equal to one.
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And now, if we set 𝑏 squared equal to 𝑐 squared minus 𝑎 squared, we’re left with 𝑥 squared over 𝑎 squared minus 𝑦 squared over 𝑏 squared is equal to one.
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And this is the equation of a horizontal or east west–opening hyperbola in standard position.
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Now, remember, the hyperbola has two asymptotes, 𝑦 is plus or minus 𝑏 over 𝑎 times 𝑥, which it approaches as 𝑥 gets large but never quite touches.
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We can see how this works with respect to our equation by solving for 𝑦 and looking at the behavior as 𝑥 approaches positive and negative ∞.
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And in fact, making 𝑦 the subject of the equation, we have 𝑦 is equal to the positive or negative square root of 𝑏 squared over 𝑎 squared 𝑥 squared minus 𝑏 squared.
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We note first that the argument of the square root is positive since 𝑥 squared is always greater than or equal to 𝑎 squared.
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And as 𝑥 approaches positive or negative ∞, the 𝑏 squared term becomes insignificant.
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Hence, 𝑦 approaches the positive or negative square root of 𝑏 squared over 𝑎 squared times 𝑥 squared, which is positive or negative 𝑏 over 𝑎 times 𝑥.
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That is the straight lines through the origin with slopes positive or negative 𝑏 over 𝑎.
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So now let’s look at an example of how to graph a hyperbola in standard position.
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Given the hyperbola defined by the equation 𝑥 squared over 36 minus 𝑦 squared over 16 is equal to one, find the vertices, the foci, and the asymptotes, and use these to graph the hyperbola.
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We’re given the equation of a hyperbola, 𝑥 squared over 36 minus 𝑦 squared over 16 is equal to one, which is in the form 𝑥 squared over 𝑎 squared minus 𝑦 squared over 𝑏 squared is equal to one, and where 𝑎 and 𝑏 are both positive.
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In our case then, 𝑎 squared is 36, and 𝑏 squared is 16.
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And so 𝑎 is six, and 𝑏 is four.
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For a hyperbola of this form, the center is at the origin zero, zero.
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If we look again at our equation, we can see that the positive term is 𝑥 squared and the negative term is 𝑦 squared.
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And this means that the major or transversed axis is on the 𝑥-axis.
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And remember that both the foci and the vertices are on the major axis, so that our hyperbola will look something like this.
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Remember that the vertices are the points on each branch which are closest to the center.
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And for a hyperbola of this type, the vertices are at negative 𝑎, zero and 𝑎, zero, so that in our case, the vertices have coordinates negative six, zero and six, zero.
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The coordinates of the foci or foci 𝐹 one and 𝐹 two are negative 𝑐, zero and 𝑐, zero, where 𝑐 squared is 𝑎 squared plus 𝑏 squared.
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So in our case, 𝑐 squared is 36 plus 16, which is 52.
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And therefore, 𝑐 which is positive is the square root of 52.
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And to one decimal place, that’s 7.2, so that our foci have coordinates negative 7.2, zero and 7.2, zero.
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The asymptotes of a hyperbola in standard position are the lines 𝑦 is 𝑏 over 𝑎 times 𝑥 and 𝑦 is negative 𝑏 over 𝑎 𝑥.
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So if we call our asymptotes 𝐴 one and 𝐴 two with 𝑎 equal to six and 𝑏 equal to four, we have 𝐴 one equal to four over six 𝑥, which is two over three 𝑥.
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And our second asymptote is negative two over three times 𝑥.
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And now we have all the information we need to graph our hyperbola.
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We have the vertices with coordinates negative six, zero and six, zero; the foci with coordinates negative 7.2, zero and 7.2, zero; and our two asymptotes.
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That’s 𝑦 is two over three 𝑥 and negative two over three 𝑥.
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Now let’s look at an example of how we find the equation of a hyperbola from its asymptotes in a real-world context.
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A gardener wants to grow a hedge in the shape of a hyperbola with a fountain at its center.
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At its closest, the hedge will be a distance of 20 yards from the fountain.
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The fountain is at the origin of the coordinate system with units in yards, and the hyperbola traced by the hedge has asymptotes 𝑦 is three over four 𝑥 and 𝑦 is negative three over four 𝑥.
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Find the equation of the hyperbola.
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To answer this question, let’s begin by sketching our hyperbola using the information given.
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We’re told that the fountain is at the center, but also that the fountain is the origin of the coordinate system.
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The hedge will be the hyperbola’s branches, which will approach the asymptotes 𝑦 is three over four 𝑥 and negative three over four 𝑥 as 𝑥 gets larger in the positive and negative directions.
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Our question says, at its closest, the hedge will be a distance of 20 yards from the fountain.
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And since the fountain is the origin of our coordinate system, we can take the points with coordinates 20, zero and negative 20, zero as our vertices.
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This means that we’ve opted for an east west–opening hyperbola.
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And with the center at the origin, this has an equation of the form 𝑥 squared over 𝑎 squared minus 𝑦 squared over 𝑏 squared is equal to one.
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And we know that for an east west–opening hyperbola, the asymptotes are 𝑦 is positive or negative 𝑏 over 𝑎 times 𝑥.
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We’re given that these are three over four and negative three over four 𝑥.
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And so it’s possible that 𝑏 is equal to three and 𝑎 is equal to four.
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Our asymptotes are the lines 𝑦 is equal to positive or negative three over four 𝑥, which are the lines with the slopes positive and negative three over four.
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However, our fractions could’ve been simplified, and 𝑎 and 𝑏 could actually be larger than these numbers.
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However, we do know somewhere else where 𝑎 appears.
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And that’s on the transverse or major axis of the hyperbola.
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We know that the distance from the center of the hyperbola to the vertex is 𝑎 since the vertices of such a hyperbola have coordinates plus or minus 𝑎, zero.
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So in our case, 𝑎 is equal to 20.
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And we know that our positive slope is 𝑏 over 𝑎, and that’s three over four.
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And since 𝑎 is 20, that means three over four is equal to 𝑏 over 20.
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If we cross multiply, we have 20 times three over four is equal to 𝑏.
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And since four divides 20 five times, our 𝑏 is equal to 15.
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And now, since we have the value of 𝑏 and the value of 𝑎, we can substitute these into our hyperbola equation where 𝑏 squared is 225 and 𝑎 squared is 400.
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And our hyperbola is 𝑥 squared over 400 minus 𝑦 squared over 225 is equal to one, so that our hedge is in the shape of a hyperbola with equation 𝑥 squared over 400 minus 𝑦 squared over 225 is equal to one.
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Now, so far, we’ve been looking at hyperbolas which were centered at the origin.
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But of course, they could be centered anywhere in the 𝑥𝑦-plane.
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If we translate our hyperbola in the plane so that its center has coordinates ℎ, 𝑘, our translated hyperbola has the equation 𝑥 minus ℎ squared over 𝑎 squared minus 𝑦 minus 𝑘 squared over 𝑏 squared is equal to one.
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In this form, since the 𝑥-term is positive, the major axis is parallel to the 𝑥-axis.
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And remember, the major axis contains the vertices and the foci.
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And this is an east west–opening hyperbola or a horizontal hyperbola.
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It’s worth noting that if the 𝑦-term was positive, then our hyperbola would look like this with the transverse or major axis parallel to the 𝑦-axis.
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And this is called a north south–opening or a vertical hyperbola.
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So now let’s see how this translation of the center of the hyperbola works in practice.
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Suppose that we model an object’s trajectory in the solar system by a hyperbolic path in the coordinate plane, with its origin at the sun and its units in astronomical units, Au.
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The 𝑥-axis is a line of symmetry of this hyperbola.
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The object enters in the direction of 𝑦 is three 𝑥 minus nine, leaves in the direction of 𝑦 is negative three 𝑥 plus nine, and passes within one Au of the sun at its closest point.
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Using the equations of the asymptotes, find the equation of the object’s path.
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We’re told that the object’s path is a hyperbola, that it enters in the direction of 𝑦 is three 𝑥 minus nine and leaves in the direction of 𝑦 is negative three 𝑥 plus nine.
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So we can assume that these two equations are the equations of the asymptotes of the hyperbola.
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In both of these equations, when 𝑦 is equal to zero, we find that 𝑥 is equal to three, so that the lines intersect at the point with coordinates three, zero.
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Our first line has a slope of three and a 𝑦-intercept at negative nine.
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And our second asymptote has a slope of negative three and a 𝑦-intercept of plus nine.
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We know that our object passes within one Au of the sun at its closest point.
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This means that one of our vertices must have coordinates one, zero, so that our hyperbola looks something like this.
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And we’re told that the 𝑥-axis is one of our lines of symmetry.
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If we take this is as our major or transverse axis, then our hyperbola is horizontal or east west–opening.
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And we’re asked to use the equations of the asymptotes to find the equation of the object’s path.
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Now we’ve established that the center of the hyperbola is at the point three, zero.
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So this is a hyperbola in standard position that’s been shifted three units in the positive 𝑥-direction, where a unit here is an astronomical unit.
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Now we know that a hyperbola with center ℎ, 𝑘 and with this orientation has the equation 𝑥 minus ℎ squared over 𝑎 squared minus 𝑦 minus 𝑘 squared over 𝑏 squared is equal to one.
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And we concede that in our case, ℎ is equal to three and 𝑘 is equal to zero.
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So that our equation will have the form 𝑥 minus three squared over 𝑎 squared minus 𝑦 squared over 𝑏 squared is equal to one.
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And all we need to do now is to find the values of 𝑎 and 𝑏.
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We can do this by using our knowledge of hyperbolas.
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Now we know that for any east west–opening hyperbola, the slope of the asymptotes is positive and negative 𝑏 over 𝑎, where 𝑎 and 𝑏 are both positive.
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In our case, 𝑏 over 𝑎 is equal to three, so that 𝑏 is equal to three 𝑎.
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Just to make some room, I’ll move things round a little bit.
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We have 𝑏 equals three 𝑎 and 𝑥 minus three squared over 𝑎 squared minus 𝑦 squared over 𝑏 squared is equal to one.
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So now if we substitute 𝑏 is equal to three 𝑎 into our equation, we have 𝑥 minus three squared over 𝑎 squared minus 𝑦 squared over nine 𝑎 squared is equal to one.
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Now we know that the hyperbola passes through the point one, zero because this is our vertex.
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So we know that when 𝑥 is equal to one, 𝑦 is equal to zero.
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And if we substitute these into our equation, we have negative two squared over 𝑎 squared is equal to one, so that 𝑎 squared is equal to four, and 𝑎 is equal to two since 𝑎 is positive.
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And remember that 𝑏 is equal to three times 𝑎, so 𝑏 is equal to six.
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And the equation of the object’s path is 𝑥 minus three squared over four minus 𝑦 squared over 36 is equal to one.
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Let’s conclude our look at the equation of a hyperbola by recapping some key points.
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A hyperbola in standard position has its center at zero, zero.
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The orientation of the hyperbola is determined by whether it’s the 𝑥 squared or the 𝑦 squared term that has the positive sign.
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A hyperbola with center ℎ, 𝑘 has been shifted or translated in the plane ℎ units in the 𝑥-direction and 𝑘 units in the 𝑦-direction.
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The vertices are the two points which are the closest to the center on the branches.
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The foci are the points a distance 𝑐 from the center such that for any point on a branch of the hyperbola, the difference in the distances from the point to the two foci is the constant two 𝑎.
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The asymptotes are the intersecting lines through the center of the hyperbola, which the branches of the hyperbola approach as 𝑥 gets large but never touch.