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Given the equation 4x^2 + 9y^2 = 36 find: a) The Center b) Length of the Major...
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The standard equation of an ellipse, with its center at `(x_1,y_1)` is:
`(x-x_1)^2/a^2 + (y-y_1)^2/b^2 = 1`
The lengths of major and minor axes are given by 2a and 2b respectively.
The distance from center to foci is ae where e is the eccentricity of the ellipse, given by ae, where,
The given equation is
`4x^2 + 9y^2 = 36`
Dividing both sides by 36,
`x^2/9 + y^2/4 = 1`
`rArr (x-0)^2/3^2+(y-0)^2/2^2 = 1`
A) This is the standard equation of an ellipse whose center is at (0,0).
B) The length of its major axis is 2*3=6 units.
C) The length of its minor axis is 2*2 = 4 units.
D) In order to calculate the distance from center to foci, the eccentricity, e has to be calculated first.
`e^2=1-(4)^2/(6)^2 = 0.55556`
`rArr e = 0.745356`
`rArr ae = 3*0.745356 = 2.2361` units.
Hence the distance from center to foci is 2.2361 units.
Posted by llltkl on June 28, 2013 at 1:08 AM (Answer #1)
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High School Teacher
Another way to find the distance from the center to the foci (for an ellipse) is to use `c^2=a^2-b^2` where c is the required length. In this problem we have a=3 and b=2 so `c^2=9-4 ==> c=sqrt(5)` .
For hyperbolas the formula is `c^2=a^2+b^2` .
Posted by embizze on June 28, 2013 at 2:55 AM (Reply #1)
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