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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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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted June 28, 2013 at 12:22 AM via web

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Given the equation 4x^2 + 9y^2 = 36 find:

a)  The Center

b)  Length of the Major Axis

c)  Length of the Minor Axis

d)  Distance from C to foci

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llltkl | College Teacher | (Level 3) Valedictorian

Posted June 28, 2013 at 1:08 AM (Answer #1)

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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,

`e^2=1-(2b)^2/(2a)^2`

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.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 28, 2013 at 2:55 AM (Reply #1)

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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` .

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