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hammer [34]
3 years ago
11

ASAP Help me please due in the morning

Mathematics
1 answer:
dybincka [34]3 years ago
3 0

                                so are you supose to like move the lines in a diffrent angle?

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An ellipse has vertices at (0, #17) and foci at (0, ±15). Write the equation of the ellipse in standard form. Graph the ellipse.
Lunna [17]

<u>ANSWER</u>

\frac{ {x}^{2} }{ 64 }  +  \frac{ {y}^{2} }{ 289 }  = 1

See attachment for the graph

<u>EXPLANATION</u>

The standard equation of the vertical ellipse with center at the origin is given by

\frac{ {x}^{2} }{ {b}^{2} }  +  \frac{ {y}^{2} }{ {a}^{2} }  = 1

where

{a}^{2}  \:  >  \:  {b}^{2}

The ellipse has its vertices at (0,±17).

This implies that:a=±17 or a²=289

The foci are located at (0,±15).

This implies that:c=±15 or c²=225

We use the following relation to find the value of b²

{a}^{2}  -  {b}^{2}  =  {c}^{2}

\implies \: 289  -  {b}^{2}  =  225

-  {b}^{2}  =  225 - 289

-  {b}^{2}  =  - 64

{b}^{2}  = 64

We substitute into the formula for the standard equation to get:

\frac{ {x}^{2} }{ 64 }  +  \frac{ {y}^{2} }{ 289 }  = 1

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