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Given the ellipse 3x^2 + 5y^2 -12x -50y + 62 = 0 find: a)  The Center C b)  The...

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

Posted June 29, 2013 at 2:11 AM via web

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Given the ellipse 3x^2 + 5y^2 -12x -50y + 62 = 0 find:

a)  The Center C

b)  The length of the major axis

c)  The length of the minor axis

d)  The distance from C to foci

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mjripalda | High School Teacher | (Level 1) Senior Educator

Posted June 29, 2013 at 3:16 AM (Answer #1)

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`3x^2+5y^2-12x-50y+62=0`

(a) To determine its center, the equation must be transformed to standard form.

Note that the standard form of ellipse are:

`(x-h)^2/a^2+(y-k)^2/b^2=1 `     and   `(x-h)^2/b^2+(y-k)^2/a^2=1`

where (h,k) is the center and a^2 should always be greater than b^2.

To express the given equation in that form , group the terms with x together. Do the same with y.

`(3x^2-12x)+(5y^2-50y)+62=0`

In order that the x^2 to have a coefficient of 1, factor out the 3 in the first group. And for the second group, factor out 5 so that y^2 would have a coefficient of 1 too.

`3(x^2-4x)+5(y^2-10y)+62=0`

Next, apply completing the square method for each group.

For the first group:

`3(x^2-4x)=3(x^2-4x+4)-12=3(x-2)^2-12`

For the second group:

`5(y^2-10y)=5(y^2-10y+25)-125=5(y-5)^2-125`

So the equation becomes:

`3(x-2)^2-12+5(y-5)^2-125+62=0`

`3(x-2)^2+5(y-5)^2-75=0`

Next, isolate the constant. To do so, add both sides by 75.

`3(x-2)^2+5(y-5)^2-75+75=0+75`

`3(x-2)^2+5(y-5)^2=75`

To have 1 at the right side of the equation, divide both sides by 75.

`(3(x-2)^2+5(y-5)^2)/75=75/75`

`(3(x-2)^2)/75+(5(y-5)^2)/75=1`

`(x-2)^2/25+(y-5)^2/15=1`

This is the standard form of the given ellipse equation. 

Hence, the center of the ellipse is `(2,5)` .

(b) To determine the length of the major axis, consider the large denominator which is 25.

This means that:

`a^2=25`

And the value of a is:

`sqrt(a^2)=+-sqrt25`

`a=+-5`

Since a represents the length of semi-major axis, take only the positive value.

`a=5`

And then, multiply it by 2 to get the whole length of the major axis.

length of major axis `=2a=2*5=10`

Hence, the length of ellipse major axis is 10 units.

(c) For the length of minor axis, consider the smaller denominator in the standard form of the given equation which is 15.

So,

`b^2=15`

And the value of b is:

`sqrt(b^2)=+-sqrt15`

`b=+-sqrt15`

Since b represents the length of semi-minor axis, take the positive value. And multiply it by 2 to get the whole length of minor axis.

length of minor axis`=2b=2*sqrt15=2sqrt15`

Hence, the length of the minor axis is `2sqrt15` units.

(d) To solve for the distance of the foci from the center of the ellipse, apply the formula:

`c=sqrt(a^2-b^2)`

Plug-in the values of a^2 and b^2.

`c=sqrt(25-15)=sqrt10`

Hence, the distance of each focus from the center of the ellipse is `sqrt10` units.

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