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Given the ellipse 25x^2 + 5y^2 + 50x - 20y + 20 = 0 find: a)  The Center C b)  The...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted June 30, 2013 at 2:40 AM via web

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Given the ellipse 25x^2 + 5y^2 + 50x - 20y + 20 = 0 find:

a)  The Center C

b)  The Length of the Major Axis

c)  The Length of the Minor Axis

d)  Distance from C to foci

 

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llltkl | College Teacher | Valedictorian

Posted June 30, 2013 at 4:05 AM (Answer #1)

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The standard forms of horizontal and vertical ellipses are respectively:

`(x-h)^2/a^2+(y-k)^2/b^2=1` and` (x-h)^2/b^2+(y-k)^2/a^2=1` where (h,k) is the center, a is semi major axis, b is semi minor axis.

The given equation is `25x^2 + 5y^2 + 50x - 20y + 20 = 0`

Rewrite the equation in the form:

 

`25x^2 + 50x+25+5y^2 -20y+20=-20+25+20 `

`rArr 25(x+1)^2+5(y-2)^2=25`

 

Dividing both sides by 25

`(x+1)^2/1+(y-2)^2/5=1`

This is the standard equation of a vertical ellipse.

1) Therefore,the Center C is (-1,2)

2) To determine the length of the major axis, consider the larger denominator which is 5.

So, `a^2=5`

`rArr` `a=sqrt5`

length of major axis =`2a=2*sqrt5=2sqrt5` units.

3) To determine the length of the minor axis, consider the smaller denominator which is 1.

So, `b^2=1`

`rArr` b=1

length of the minor axis = 2b=2*1=2 units.

4) To find the distance, c of the foci from the center of the ellipse, apply the formula:

`c=sqrt(a^2-b^2)`

Plugging in the values of `a^2` and `b^2`

`C=sqrt(5-1)` =2 units

Hence, the distance of each focus from the center of the ellipse is 2 units.

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