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# Help asap.

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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
+1014
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By Ellipse do you mean a Circle

Mar 20, 2019
#2
+38
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No, like the oval shape

Dahdahda  Mar 20, 2019
edited by Dahdahda  Mar 20, 2019
edited by Dahdahda  Mar 20, 2019
#3
+1014
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Okay Well it look like you are asking for the botom of the Oval right

Nickolas  Mar 20, 2019
#4
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I need d in the second focus of the ellipse

Dahdahda  Mar 20, 2019
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+110153
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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
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+110153
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This shoud be helpful I think

Mar 20, 2019
#7
+38
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Cool. I watched it, but I still don’t even know how to do this.

Dahdahda  Mar 20, 2019
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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