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An ellipse in the first quadrant is tangent to both the x-axis and y-axis. One focus is at  (3, 7) and the other focus is at (d, 7). Compute d.

 Mar 20, 2019
 #1
avatar+1014 
-5

By Ellipse do you mean a Circle 

 Mar 20, 2019
 #2
avatar+38 
+2

No, like the oval shape

Dahdahda  Mar 20, 2019
edited by Dahdahda  Mar 20, 2019
edited by Dahdahda  Mar 20, 2019
 #3
avatar+1014 
-5

Okay Well it look like you are asking for the botom of the Oval right

 

If not please tell me all you know about this..

I sincerly would love to help you.

Nickolas  Mar 20, 2019
 #4
avatar+38 
+1

I need d in the second focus of the ellipse

Dahdahda  Mar 20, 2019
 #5
avatar+110153 
+2

Nickolas, this is way over your head :)

 

Dah, a circle is actually a special case of an ellipse.

 

A circle is an ellipse where for foci are the same point.

 Mar 20, 2019
edited by Melody  Mar 20, 2019
 #6
avatar+110153 
+2

This shoud be helpful I think

 

https://www.youtube.com/watch?v=llJuk41PNIg

 Mar 20, 2019
 #7
avatar+38 
0

Cool. I watched it, but I still don’t even know how to do this.

Dahdahda  Mar 20, 2019
 #8
avatar+395 
+1

If the first foci is \((3,7) \) and the second is \((d, 7)\), then \(C = \) \((\frac{d+3}{2}, 7)\). \(C\) is the center of the ellipse. The point where the ellipse touches the x-axis is \((\frac{d+3}{2}, 0)\).

 

We know that \(P\)(any points on the ellipse) = \(T\), so

\(2 \sqrt{\left( \frac{d - 3}{2} \right)^2 + 7^2} = d + 3\)  

\(\sqrt{\left( \frac{d - 3}{2} \right)^2(4) + 7^2(4)} = d + 3\)

\(\sqrt{(d - 3)^2 + 196} = d + 3\)

 

Then, by squaring both sides, we get 

\((d - 3)^2 + 196 = d^2 + 6d + 9\)

\(d^2-6d+9+196=d^2+6d+9\)

\(12d=196\)

\(d=\frac{49}{3}\)

 

- Daisy

 Mar 20, 2019

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