How do you solve this ellipse?

Question

The center is (4, -1), the focus is (1,-1), and the ellipse passes through (8,0)

Please explain, thoroughly, what needs to be done in order to obtain all those other important points, and the standard equation of this ellipse.

all I'm giving you is the center of the ellipse which would be the point at which the major and minor axis intersect

Answer 1

We should first address this problem by putting it into the standard form of the horizontal ellipse, which is

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

Center: (h,k)
Foci: (h+c,k), (h-c,k)
Endpoints of Horizontal Axis: (h+a,k), (h-a,k)
Endpoints of Vertical Axis: (h,k+b), (h,k-b)
Note that a^2=b^2+c^2.

Now we apply the known information:

Center: (4,-1) >>> h=4 , k=-1
Focus: (1,-1) >>> c=3,-3
>>> Focus: (7,-1) --- (h+-c,k)
Point: (8,0)

Therefore (in standard form),
(x-4)^2/a^2 + [y-(-1)]^2/b^2 = 1
(x-4)^2/a^2 + (y+1)^2/b^2 = 1

Next we use the property PF'1+PF'2 = 2a.

sqrt{[8-1]^2+[0-(-1)]^2} + sqrt{[8-7]^2+[0-(-1)]^2} = 2a
6sqrt(2) = 2a
a = 3sqrt(2)

Applying a^2=b^2+c^2, we have

[3sqrt(2)]^2 = b^2 + (+-3)^2
18 = b^2+9
b^2=9
b = +-3

Substituting into the standard equation,

(x-4)^2/[3sqrt(2)]^2 + (y+1)^2/(+-3)^2 = 1

(x-4)^2/18 + (y+1)^2/9 = 1

Center: (4,-1)
Foci: (1,-1), (7,-1)
Endpoints of horizontal axis (h+-a,k): [4+-3sqrt(2),-1]
Endpoints of vertical axis (h,k+-b): (4,2), (4,-4)

Answer 2

Find the equation of the ellipse.

The center is (4, -1), the focus is (1,-1), and the ellipse passes through (8, 0).

The major axis runs thru the center and the focus, so the major axis is horizontal.

The distance from the center to a focus is c.

c = 4 - 1 = 3
c² = 9

The other focus is an equal distance away on the other side of the center.

4 + 3 = 7

The other focus is (7, -1).

The distance from one focus to any point on the ellipse to the other focus is 2a.

2a = √[(1-8)² + (-1-0)²] + √[(8-7)² + (0- -1)²]
2a = √(7² + 1²) + √(1² + 1²) = √50 + √2 = 5√2 + √2 = 6√2

a = 3√2
a² = 18

b² = a² - c² = 18 - 9 = 9

The equation of the ellipse is:

(x - h)²/a² + (y - k)²/b² = 1

(x - 4)²/18 + (y + 1)²/9 = 1

Answer 3

Since the center and 'one' of the foci are aligned horizontally, this means that the principal axis is parallel to the x-axis. That means the equation is
(x-4)^2/a^2 + (y+1)^2/b^2 = 1, a>b
taking into consideration the location of the center.

Now, the distance bet the center and the focus is c = 3.
(Where a^2 = b^2 + c^2).

Thus, (x-4)^2/(b^2 +9) + (y+1)^2/b^2 =1.
Next, If you are given any point on a curve, this simply means the point satisfies the equation. You can plug the point into the equation.
Giving us 16/(b^2+9) + 1/b^2 = 1.
This might seem unwieldy but it is actually an equation in quadratic form if arranged. For simplification let N = b^2.
==> 16N + N + 9 = N(N+9)
==> N^2 -8N -9=0 (You can factor this now, right.)
Note: one solution for N=b^2 is negative and thus extraneous.
Finally, b = 3 and a = 3*sqrt2 .And I have given you the answer. ;)
Hope it clears things.

Answer 4

An ellipse has *two* foci.

Do you mean the foci are (4,-1) and (1,-1) or are we dealing with another geometric form?