# Thread: Finding equation of ellipse

1. ## Finding equation of ellipse

The major axis of ellipse has end points (8,4) and (-4,4). If the ellipse passes through the origin, what is the equation of the ellipse?

My solution:

End points (8,4) and (-4,4)
so 2a = 12, a =6

therefore:
x²/a² + y²/b² =1

x²/6² + y²/b² =1

Passes through origin, put (0,0) in x and y:

0²/6² + 0²/b² =1
b² = 0
b = 0

So the equation for the ellipse is:

x²/6² + y² =1

Did I do this right? Is it really x²/6² + y² =1 ?

Thanks.

2. Originally Posted by shenton
The major axis of ellipse has end points (8,4) and (-4,4). If the ellipse passes through the origin, what is the equation of the ellipse?

My solution:

End points (8,4) and (-4,4)
so 2a = 12, a =6

therefore:
x²/a² + y²/b² =1

x²/6² + y²/b² =1

Passes through origin, put (0,0) in x and y:

0²/6² + 0²/b² =1
b² = 0
b = 0

So the equation for the ellipse is:

x²/6² + y² =1

Did I do this right? Is it really x²/6² + y² =1 ?

Thanks.
Since when does 0/b^2 = 1 imply b = 0 and at what point does 1/0 = 1?

Let's try this again...

Yes, a = 6.

But the equation for the ellipse would be:
(x - h)^2/36 + (y - k)^2/b^2 = 1

Now, we know the center of the ellipse is the point (2, 4) since this is the midpoint of the line segment containing the two vertices. Thus h = 2 and k = 4:
(x - 2)^2/36 + (y - 4)^2/b^2 = 1

Now put in the origin:
4/36 + 16/b^2 = 1

1/9 + 16/b^2 = 1

(1/9)b^2 + 16 = b^2

(1/9 - 1)b^2 = -16

-(8/9)b^2 = -16

b^2 = 18

b = sqrt{18}

So:
(x - 2)^2/36 + (y - 4)^2/18 = 1

-Dan

3. Thanks. I knew my solution looks too bad to be right.