# 12.1 The ellipse  (Page 5/16)

 Page 5 / 16

What is the standard form equation of the ellipse that has vertices $\text{\hspace{0.17em}}\left(-3,3\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,3\right)\text{\hspace{0.17em}}$ and foci $\text{\hspace{0.17em}}\left(1-2\sqrt{3},3\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1+2\sqrt{3},3\right)?$

$\frac{{\left(x-1\right)}^{2}}{16}+\frac{{\left(y-3\right)}^{2}}{4}=1$

## Graphing ellipses centered at the origin

Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form for horizontal ellipses and for vertical ellipses.

Given the standard form of an equation for an ellipse centered at $\text{\hspace{0.17em}}\left(0,0\right),$ sketch the graph.

1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
1. If the equation is in the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a>b,\text{\hspace{0.17em}}$ then
• the major axis is the x -axis
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(±a,0\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(0,±b\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(±c,0\right)$
2. If the equation is in the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,$ where $\text{\hspace{0.17em}}a>b,\text{\hspace{0.17em}}$ then
• the major axis is the y -axis
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(0,±a\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(±b,0\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(0,±c\right)$
2. Solve for $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.$
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

## Graphing an ellipse centered at the origin

Graph the ellipse given by the equation, $\text{\hspace{0.17em}}\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, and foci.

First, we determine the position of the major axis. Because $\text{\hspace{0.17em}}25>9,$ the major axis is on the y -axis. Therefore, the equation is in the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,$ where $\text{\hspace{0.17em}}{b}^{2}=9\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{a}^{2}=25.\text{\hspace{0.17em}}$ It follows that:

• the center of the ellipse is $\text{\hspace{0.17em}}\left(0,0\right)$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(0,±a\right)=\left(0,±\sqrt{25}\right)=\left(0,±5\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(±b,0\right)=\left(±\sqrt{9},0\right)=\left(±3,0\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(0,±c\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,$ we have:

$\begin{array}{l}c=±\sqrt{{a}^{2}-{b}^{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{25-9}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{16}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±4\hfill \end{array}$

Therefore, the coordinates of the foci are $\text{\hspace{0.17em}}\left(0,±4\right).$

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See [link] .

Graph the ellipse given by the equation $\text{\hspace{0.17em}}\frac{{x}^{2}}{36}+\frac{{y}^{2}}{4}=1.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, and foci.

center: $\text{\hspace{0.17em}}\left(0,0\right);\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(±6,0\right);\text{\hspace{0.17em}}$ co-vertices: $\text{\hspace{0.17em}}\left(0,±2\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(±4\sqrt{2},0\right)$

## Graphing an ellipse centered at the origin from an equation not in standard form

Graph the ellipse given by the equation $\text{\hspace{0.17em}}4{x}^{2}+25{y}^{2}=100.\text{\hspace{0.17em}}$ Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

First, use algebra to rewrite the equation in standard form.

Next, we determine the position of the major axis. Because $\text{\hspace{0.17em}}25>4,\text{\hspace{0.17em}}$ the major axis is on the x -axis. Therefore, the equation is in the form $\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{a}^{2}=25\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{b}^{2}=4.\text{\hspace{0.17em}}$ It follows that:

• the center of the ellipse is $\text{\hspace{0.17em}}\left(0,0\right)$
• the coordinates of the vertices are $\text{\hspace{0.17em}}\left(±a,0\right)=\left(±\sqrt{25},0\right)=\left(±5,0\right)$
• the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(0,±b\right)=\left(0,±\sqrt{4}\right)=\left(0,±2\right)$
• the coordinates of the foci are $\text{\hspace{0.17em}}\left(±c,0\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,\text{\hspace{0.17em}}$ we have:

$\begin{array}{l}c=±\sqrt{{a}^{2}-{b}^{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{25-4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{21}\hfill \end{array}$

Therefore the coordinates of the foci are $\text{\hspace{0.17em}}\left(±\sqrt{21},0\right).$

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
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nothing up todat yet
Miranda
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jai
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Miranda Drice
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Miranda
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Miranda
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I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
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Propessor
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Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
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jai
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Propessor
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algebra is a branch of the mathematics to calculate expressions follow.
Miranda
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Jeffrey
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Miranda
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Miranda
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Steve
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Miranda
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Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
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Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa
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(x+x)3=?
6x
Obed
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Kishu
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what is the function of sine with respect of cosine , graphically
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Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls