# Given the elipse 4x^2 + 9y^2 + 32x - 144y + 604 = 0 Find: a) The Center b) Length of Major Axis c) Length of Minor Axis d) Distance from C to foci

llltkl | College Teacher | (Level 3) Valedictorian

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Given equation of ellipse is:

`4x^2 + 9y^2 + 32x - 144y + 604 = 0`

Rewrite the equation in the form:

`4x^2+32x+64+9y^2-144y+576=-604+64+576`

`rArr 4(x^2+8x+16)+9(y^2-16y+64)=36`

`rArr 4(x+4)^2+9(y-8)^2=36`

Divide both sides by 36,

`(x+4)^2/3^2+(y-8)^2/2^2=1`

This is the standard equation of a horizontal ellipse.

1. Its center C is at (-4,8).

2. To determine the length of the major axis, consider the larger denominator which is 9.

So, `a^2=3^2`

`rArr a=3`

length of major axis = 2a=2*3=6 units.

3. Similarly,  `b^2=2^2`

`rArr b=2`

length of minor axis = 2b=2*2=4 units.

4. To find the distance 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(9-4) =sqrt5`  units.

Hence, the distance of each focus from the center of the ellipse is `sqrt5` units.

Sources:

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