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

durbanville | High School Teacher | (Level 2) Educator Emeritus

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Please post the remaining sections (c) and d)) as another question because the eNotes rules do not allow educators to answer more than one question at a time. A closely-related question can be answered where appropriate; hence, I was able to answer a) and b).

durbanville | High School Teacher | (Level 2) Educator Emeritus

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

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

Complete the square. Remember to reduce the co-efficient of x and y to 1 if necessary:

`8(x^2+10x) +(y^2-6y) = -193`

Using 10x and -6y: `(10/2)^2= 5^2`     and    `(-6/2)^2= (-3)^2` . Note the unsquared result on the LHS (left hand side) and the squared result on the RHS as the LHS will still be squared in the bracket! Note also to include the 8 on the RHS when adding the squared results:

`therefore 8(x+5)^2 +(y-3)^2= -193 +8(25) +9`

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

We now have the center: `C=(-5;3)`

To find the vertices or major axis, find a from `b^2+c^2=a^2` :

Reduce the RHS to =1:

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

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

a is always the larger denominator:

`therefore b^2+c^2 = a^2`  becomes

`2 + c^2 = 16` . So we know that `a=4`

The vertices are a=4 units above and below the center (-5;3):

`therefore = (-5;-1) and (-5; 7)`

Therefore Ans:

C(-5;3) and major axis (-5;-1) (-5;7)

kristenmariebieber | Student, Grade 10 | (Level 1) Valedictorian

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Thank you for responding to the question.  However, there are

two parts to the question that were not answered.

What is the length of the Minor Axis?

What is the distance from C to foci?