# Given the ellipse 8x^2 + y^2 + 80x - 6y + 193 = 0 find: c) The Length of the Minor Axis d) Distance from C to foci

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### 2 Answers

We can write the ellipse in the form:

`((x-p)^2)/a^2+((y-q)^2)/b^2=1`

Where:

(p,q) is the center of the ellipse,

a is half of the major axis or the distance from the center to the intersection of the ellipse with the major axis and

b is half of the minor axis or the distance from the center to the intersection of the ellipse with the major axis

Grouping to find the squares, add 16 to each side:

`8(x^2+10x+25)+(y^2-6y+9)` =16

`8(x+5)^2+(y-3)^2=16`

Dividing by 16, the equation of the ellipse is:

`((x+5)^2)/2+((y-3)^2)/16=1`

Where:

`a=sqrt16=4`

`b=sqrt2`

Center: (-5,3)

Because the major axis is parellel to y, this is a vertical ellipse.

To find the distance from the center to foci we use the definition of an ellipse for a point at the intersection with the minor axis (we will call it point B): the sum of the distances from the foci to B is equal to 2a. Since B is equidistant to each focus, the from a focus to B is a=4.

We also know that the distance from the center to point B is equal to b=`sqrt2`

Since the triangle FCB is a right triangle with hipothenuse 4 and we also know one side to be `sqrt2` , we can use the Pythagorean Theorem to find the third side:

`FC=sqrt(4^2-2)=sqrt14=3.74`

**Length of the minor axis = `2sqrt2=2.8` **

**Distnace from the center to foci = 3.74**

Given equation of ellipse is:

`8x^2 + y^2 + 80x - 6y + 193 = 0`

Rewrite the equation in the form:

`8x^2+80x+200+y^2-6y+9=-193+200+9`

`rArr 8(x^2+10x+25)+y^2-6y+9=16`

`rArr 8(x+5)^2+(y-3)^2=16`

Divide both sides by 16 to get:

`(x+5)^2/(sqrt2)^2+(y-3)^2/(4)^2=1`

This is the standard equation of a vertical ellipse.

c) To determine the length of the minor axis, consider the smaller denominator which is 2.

So, `b^2=2`

`rArr b=sqrt2`

**length of minor axis** = `2b` =`2*sqrt2` =`2sqrt2` units.

d) 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(16-2)=sqrt14` units.

**Hence, the distance of each focus from the center of the ellipse is** `sqrt14` ** units.**

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