A parabola is a unique type of curve that forms part of a conic section, a curve obtained as the intersection of the surface of a cone with a plane. Depending on the orientation of the cone and the plane, a parabola can open in one of four directions: left, right, up, or down. One of the key features of a parabola is its vertex, the point where it reaches a maximum or minimum value, depending on its orientation.

To represent a parabola analytically, we can use the vertex form of its equation, which allows us to quickly identify the vertex and the direction in which the parabola opens. This form of the equation uses the coordinates of the vertex (h,\, k) and a constant a. If the parabola opens up or down, the equation takes the form y-k = a(x -h)^{2}. Here, a is a nonzero constant that influences the 'width' or 'narrowness' of the parabola. The larger the absolute value of a, the narrower the parabola. If a is positive, the parabola opens upwards. If a is negative, the parabola opens downwards.

A parabola that opens to the right or left can be represented by the equation x-h = a(y - k)^{2}, where (h,\, k) is the vertex of the parabola and a is a nonzero constant. The value of a determines the width or "spread" of the parabola and the direction in which it opens. If a is positive, the parabola opens to the right, while if a is negative, it opens to the left. The larger the absolute value of a, the narrower the parabola, and the smaller the absolute value of a, the wider the parabola. The vertex form of the parabola's equation offers a quick way to identify the vertex and the direction of opening, providing key information for graphing or analyzing the parabola.

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Green Equation: y-1 = 2(x + 1)^{2},\, a = 2, vertex (-1,\,1) and opens up.

Blue Equation: y + 1 = -\dfrac{1}{2}(x-1)^{2},\, a = -\dfrac{1}{2}, vertex (1,\,-1) and opens down.

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Purple Equation: x-2 = \dfrac{1}{3}(y - 1)^{2},\, a = \dfrac{1}{3}, vertex (2,\,1) and opens right.

Blue Equation: x - 1 = -1(y - 1)^{2},\, a = -1, vertex (1,\,1) and opens left.

Examples

Example 1

Let's consider the equation of a parabola given as y = 2(x - 3)^{2} + 1. Our task is to find the vertex and the direction in which the parabola opens.

Worked Solution

Create a strategy

To determine the vertex of the parabola, we can directly read off the values of h and k from the given equation, as it is already in the vertex form y = a(x - h)^{2} + k. The vertex is (h,\, k).

To determine the direction in which the parabola opens, we look at the coefficient a of (x - h)^{2}. If a is positive, the parabola opens upwards, if a is negative, it opens downwards.

Apply the idea

Here, the given equation is in the form y = a(x - h)^{2} + k, where a = 2,\, h = 3,\, and k = 1.

So, the vertex of the parabola is (h,\, k) = (3,\, 1).

The coefficient a is positive (2), so the parabola opens upwards.

Reflect and check

The vertex of the parabola is (3,\, 1), and it opens upwards. We can check this by graphing the given equation and confirming that the graphed parabola has its vertex at (3,\, 1) and opens upwards.

Example 2

Write the equation of a parabola given the vertex at (-2,\, 5) that opens downwards.

Worked Solution

Create a strategy

We can use the vertex form of a parabolic equation, y = a(x - h)^{2} + k, to write the equation. Given that the parabola opens downwards, we know that a will be negative. For simplicity, let's choose a = -1.

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle -1(x - (-2))^2 + 5	Substitute a = -1,\, h = -2 and k = 5
\displaystyle y	\displaystyle =	\displaystyle -(x + 2)^{2} + 5	Evaluate

Example 3

Consider the equation x - 3 = 2(y - 1)^{2}, identify its vertex, direction of opening, and draw its sketch.

Worked Solution

Apply the idea

From the given equation x - 3 = 2(y - 1)^{2}, we can identify the vertex (h,\, k) as (3,\, 1) since h corresponds to the constant on the x term and k corresponds to the constant in the y term.

The coefficient a is 2, which is positive, indicating that the parabola opens to the right.

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Reflect and check

The vertex of the parabola is at the point (3,\, 1), and it opens to the right. When sketching the parabola, ensure that the curve opens in the correct direction from the vertex. One can also plot additional points for accuracy, for example, by substituting y = 0and y = 2 into the equation to find the corresponding x values. With these points and the vertex, we should have a clear sketch of the parabola.

Idea summary

A parabola that opens up or down: y-k = a(x - h)^{2}

A parabola that opens to the left or right: x-h = a(y - k)^{2}

In these formulas, (h,\, k) is the vertex, and a is a non-zero constant that affects the parabola's width. The parabola opens upwards if a is positive and downwards if a is negative in the first formula. In the second formula, if a is positive, the parabola opens to the right, and if a is negative, it opens to the left.

Ellipses and their analytical representation

An ellipse is another type of curve in the family of conic sections, characterized by two focal points. Each point on the ellipse maintains a specific relationship to these foci: the sum of the distances from any point on the ellipse to the two foci is always the same, regardless of where on the ellipse the point is located. This property is what gives the ellipse its distinctive, elongated shape.

The analytic representation of an ellipse utilizes the equation \dfrac{(x-h)^{2}}{a^{2}} + \dfrac{(y-k)^{2}}{b^{2}} = 1. In this equation, (h,\, k) refers to the coordinates of the center of the ellipse, a kind of 'balance point' in the middle of the ellipse. The terms a and b represent the horizontal and vertical radii of the ellipse. These values give the lengths of the major and minor axes (the longest and shortest diameters) of the ellipse, and play a crucial role in shaping its form.

A special case of an ellipse is the circle. A circle is essentially an ellipse in which the lengths of the major and minor axes are the same, meaning that a equals b. In this case, the equation of the circle simplifies to (x-h)^{2} + (y-k)^{2} = r^{2}, where r is the radius of the circle.

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Green Ellipse Equation:

\dfrac{(x-2)^{2}}{4^{2}}+\dfrac{(y-4)^{2}}{3^{2}}=1

Purple Ellipse Equation:

\dfrac{(x-(-2))^{2}}{1^{2}}+\dfrac{(y-(-4))^{2}}{2^{2}}=1

Blue Ellipse Equation:

(x-4)^{2} + (y + 3)^{2} = 3^{2}

Examples

Example 4

Given the center (3,\,-2), horizontal radius 4, and vertical radius 4 of an ellipse, find the equation of the ellipse. Then identify the type of the ellipse (whether it is a special case of a circle or not). If it is a circle, find the radius.

Worked Solution

Apply the idea

\displaystyle \dfrac{(x-h)^{2}}{a^{2}} + \dfrac{(y-k)^{2}}{b^{2}}	\displaystyle =	\displaystyle 1	Write the general equation
\displaystyle \dfrac{(x-3)^{2}}{4^{2}} + \dfrac{(y-2)^{2}}{4^{2}}	\displaystyle =	\displaystyle 1	Substitute h=3,\,k=2,\,a^{2}=4 and b^{2}= 4

\displaystyle r	\displaystyle =	\displaystyle \sqrt{a^{2}}	Write the formula
	\displaystyle =	\displaystyle \sqrt{4^{2}}	Substitute a^{2}=4
	\displaystyle =	\displaystyle 4	Evaluate

So, the equation of the circle is (x-3)^{2} + (y+2)^{2} = 4^{2}.

Example 5

Given the equation \dfrac{(x-3)^{2}}{16} + \dfrac{(y+2)^{2}}{9} = 1.

Determine the center.

Worked Solution

Create a strategy

We can use this formula:\dfrac{(x-h)^{2}}{a^{2}} + \dfrac{(y-k)^{2}}{b^{2}} = 1 to identify the values of h,\, k,\, a,\, and b.

Apply the idea

The values are h=3,\,k=-2,\,a=4,\, and b=3.

The center is (h,\,k) which is (3,\,-2).

Determine the lengths of the major and minor axes.

Worked Solution

Apply the idea

Major axis:

\displaystyle 2a	\displaystyle =	\displaystyle 2(4)	Substitute a=4
	\displaystyle =	\displaystyle 8	Evaluate

Minor axis:

\displaystyle 2b	\displaystyle =	\displaystyle 2(3)	Substitute b=3
	\displaystyle =	\displaystyle 6	Evaluate

Identify whether it is a circle or an ellipse.

Worked Solution

Create a strategy

Check if a equals b.

Apply the idea

a=4,\,b=3,\, so a \neq b.

a is not equal to b therefore, the equation is an ellipse.

Idea summary

An ellipse is a particular type of curve that belongs to the family of conic sections, and is defined by two focal points.

The equation \dfrac{(x-h)^{2}}{a^{2}} + \dfrac{(y-k)^{2}}{b^{2}} = 1 analytically represents an ellipse, where (h,\, k) are the coordinates of the ellipse's center.

A circle is a special instance of an ellipse. In a circle, the lengths of the major and minor axes are identical.

Hyperbolas and their analytical representation

A hyperbola, a unique member of the conic sections family. Where an ellipse is a single, continuous curve and a parabola a curve opening in one direction, a hyperbola consists of two distinct curves, each a mirror image of the other. These are called the branches of the hyperbola.

Each branch of the hyperbola features a point known as a vertex, and the point that lies exactly in the middle of the vertices is known as the center of the hyperbola. The center serves as a kind of balancing point, akin to the role played by the center in an ellipse or the vertex in a parabola.

The equation representing a hyperbola varies depending on its orientation. If the hyperbola opens left and right, we use the equation \dfrac{(x-h)^{2}}{a^{2}} - \dfrac{(y-k)^{2}}{b^{2}} = 1, where (h,\, k) denotes the center of the hyperbola. On the other hand, if the hyperbola opens up and down, we use the equation \dfrac{(y-k)^{2}}{b^{2}} - \dfrac{(x-h)^{2}}{a^{2}} = 1. In these equations, a and b stand for the distances from the center to the vertices and to the lines of symmetry.

A key feature of a hyperbola is the presence of asymptotes, which are lines that the hyperbola approaches but never crosses as it extends to infinity. These asymptotes intersect at the center of the hyperbola and form a kind of boundary for the hyperbola. They're given by the equations y-k = \pm \dfrac{b}{a}(x - h), where the \pm indicates that there are two asymptotes, one positive and one negative.

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The diagram illustrates a hyperbola that opens left and right. The equation is shown as \dfrac{(x-2)^{2}}{4} - \dfrac{(y+3)^{2}}{9} = 1. It's clear that the center of the hyperbola is at (2,\, -3). The vertices are at points (0,\, -3) and (4,\, -3), and the asymptotes(in red) are given by the lines y+3 = \pm \dfrac{3}{2} (x - 2).

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Examples

Example 6

Given the equation of a hyperbola, \dfrac{(x-3)^{2}}{9} - \dfrac{(y+2)^{2}}{4} = 1.

Identify the center.

Worked Solution

Create a strategy

We can use this formula:\dfrac{(x-h)^{2}}{a^{2}} + \dfrac{(y-k)^{2}}{b^{2}} = 1 to identify the values of h,\, k,\, a,\, and b.

Apply the idea

The values are h=3,\,k=-2,\,a=3,\, and b=2.

The center is (h,\,k) which is (3,\,-2).

Identify the vertices.

Worked Solution

Create a strategy

We can use the formula: (h + a,\,k),\,(h-a,\,k).

Apply the idea

\displaystyle (h + a,\,k)	\displaystyle =	\displaystyle (3 + 3,\,-2)	Substitute h = 3,\,a=3 and k=-2
	\displaystyle =	\displaystyle (6,\,-2)	Evaluate

\displaystyle (h - a,\,k)	\displaystyle =	\displaystyle (3 - 3,\,-2)	Substitute h = 3,\,a=3 and k=-2
	\displaystyle =	\displaystyle (0,\,-2)	Evaluate

The vertices are (6,\,-2) and (0,\,-2).

Identify the asymptotes.

Worked Solution

Create a strategy

Remember that asymptotes of the hyperbola are given by the equation y-k = \pm \dfrac{b}{a}(x - h).

Apply the idea

\displaystyle y+2	\displaystyle =	\displaystyle \pm \dfrac{2}{3}(x - 3)	Substitute h=3,\,k=−2,\,a=3,\, and b=2
\displaystyle y	\displaystyle =	\displaystyle \pm \dfrac{2}{3}(x - 3)-2	Simplify

Sketch the hyperbola.

Worked Solution

Apply the idea

The center of the hyperbola is (3,\, -2), the vertices are at (0,\, -2) and (6,\, -2), and the asymptotes are given by the equations y = \dfrac{2}{3} (x - 3) - 2 and y = -\dfrac{2}{3} (x - 3) - 2. When we graph these points and lines, we get a hyperbola opening to the left and right, confirming that our solution is correct.

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Example 7

Given a hyperbola with vertices at (0,\, 3) and (0,\, -3), and with the distance from the center to the line of symmetry being 2 units.

Find the equation of the hyperbola that opens up and down.

Worked Solution

Create a strategy

We start by identifying the values of (h,\, k),\, a, and b. The center of the hyperbola (h,\, k) is the midpoint of the vertices. a is the distance from the center to a vertex, and b is the distance from the center to the line of symmetry. Once we have these values, we can plug them into the equation of a hyperbola that opens up and down: \dfrac{(y-k)^2}{a^2} - \dfrac{(x-h)^2}{b^2} = 1.

Apply the idea

Since the vertices are at (0,\, 3) and (0,\, -3), the center of the hyperbola is at the midpoint of the vertices, which is (0,\, 0). Thus, h = 0 and k = 0.

The distance from the center to a vertex is a, which is 3 units. The distance from the center to the line of symmetry is b, which is 2 units.

\displaystyle \dfrac{(y-0)^2}{3^2} - \dfrac{(x-0)^2}{2^2}	\displaystyle =	\displaystyle 1	Substitute h=0,\,k=0,\,a=3 and b=2
\displaystyle \dfrac{y^2}{9} - \dfrac{x^2}{4}	\displaystyle =	\displaystyle 1	Simplify

Determine the equations of the asymptotes.

Worked Solution

Create a strategy

Use formula for the asymptotes: y-k = \pm \dfrac{b}{a} (x - h).

Apply the idea

\displaystyle y-0	\displaystyle =	\displaystyle \pm \dfrac{2}{3} (x - 0)	Substitute h=0,\,k=0,\,a=3 and b=2
\displaystyle y	\displaystyle =	\displaystyle \pm \dfrac{2}{3}x	Simplify

Idea summary

The equation of a hyperbola depends on its orientation. If it opens left and right, the equation is \dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} =1, where (h,\, k) represents the center. If it opens up and down, the equation becomes \dfrac{(y-k)^2}{b^2} - \dfrac{(x-h)^2}{a^2} =1. Here, a and b denote distances from the center to the vertices and to the lines of symmetry, respectively.

A defining characteristic of a hyperbola is its asymptotes. They follow the equations y-k = \pm \dfrac{b}{a}(x - h), indicating that there are two asymptotes, one positive and one negative.

4.6A Conic sections

Introduction

Ideas

Parabolas and their analytical representation

Examples

Example 1

Example 2

Example 3

Ellipses and their analytical representation

Examples

Example 4

Example 5

Hyperbolas and their analytical representation

Examples

Example 6

Example 7

Outcomes

4.6.A

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