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Given the equation 4x^2 + y^2 - 48x -4y + 48 = 0 find: a) The Center C b) Length of...
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Given equation is:
`4x^2 + y^2 - 48x -4y + 48 = 0`
Rewrite the equation in the form:
`4x^2- 48x + 144 + y^2 -4y+4=-48 +144+4`
`rArr 4(x^2-12x+36) + y^2 -4y+4=100`
`rArr 4(x-6)^2 + (y-2)^2=100`
Divide both sides by 100,
`(x-6)^2/5^2 + (y-2)^2/10^2=1`
This is the standard equation of a vertical ellipse.
1. Its center C is at (6,2)
2. To determine the length of the major axis, consider the larger denominator which is 100.
length of major axis = 2a=2*10=20 units.
3. Similarly, `b^2=5^2`
length of minor axis = 2b=2*5=10 units.
4. To find the distance of the foci from the center of the ellipse, apply the formula:
Plugging in the values of `a^2` and `b^2`
c=`sqrt(100-25)` =`sqrt(75)` =`5sqrt3` units
Hence, the distance of each focus from the center of the ellipse is `5sqrt3` units.
Posted by llltkl on June 30, 2013 at 5:43 AM (Answer #1)
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