# 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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### 1 Answer

`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.**