# Given the equation 4x^2 + y^2 - 48x -4y + 48 = 0 find: a)  The Center C b)  Length of Major Axis c)  Length of Minor Axis d)  Distance from C to foci

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Given equation is:

`4x^2 + y^2 - 48x -4y + 48 = 0`

Rewrite the equation in the form:

`4x^2- 48x + 144 + y^2 -4y+4=-48 +144+4`

`rArr 4(x^2-12x+36) + y^2 -4y+4=100`

`rArr 4(x-6)^2 + (y-2)^2=100`

Divide both sides by 100,

`(x-6)^2/5^2 + (y-2)^2/10^2=1`

This is the standard equation of a vertical ellipse.

1. Its center C is at (6,2)

2. To determine the length of the major axis, consider the larger denominator which is 100.

So, `a^2=10^2`

`rArr a=10`

length of major axis = 2a=2*10=20 units.

3. Similarly,  `b^2=5^2`

`rArr b=5`

length of minor axis = 2b=2*5=10 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(100-25)` =`sqrt(75)` =`5sqrt3`  units

Hence, the distance of each focus from the center of the ellipse is `5sqrt3` units.

Sources: