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If the major aixs is vertical and has a length of 10 units, the minor axis has a length...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted June 28, 2013 at 1:57 AM via web

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If the major aixs is vertical and has a length of 10 units, the minor axis has a length of 8 units, and the Center C = (4,-2) fill in the missing denominators for the equation and determine the distance from C to the foci.

(x-4)^2 + (y+2)^2 = 1, distance from C to foci

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted June 28, 2013 at 3:22 AM (Answer #1)

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`(x-4)^2+(y+2)^2=1`

Notice that the operation between the (x-4)^2 and (y+2)^2 is addition, which indicates that the figure is an ellipse.

Since it is an ellipse and has a vertical major axis, to complete the equation, apply the formula:

`(x-h)^2/b^2+(y-k)^2/a^2=1`

where a is the length of semi-major axis and b is the length of semi-minor axis.

So to get the value of a, take half of the length of major axis which is 10.

`a= 10/2`

`a=5`

And to determine b, take half the length of minor axis which is 8.

`b=8/2`

`b=4`

Then, plug-in these values of a and b to the formula.

`(x-4)^2/4^2 + (y+2)^2/5^2=1`

Hence, the complete equation is `(x-4)^2/16+(y+2)^2/25=1` .

 

Next, to determine the distance of the focus from the center, apply the formula:

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

Plug-in a=5 and b=4.

`c=sqrt(5^2-4^2)=sqrt(25-16)=sqrt9`

`c=3`

Hence, the distance of each focus from the center C is 3 units.

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