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An ellipse has foci at (0,2) and (3,0). The ellipse has two x-intercepts, one of which is the origin. What is the other x-intercept? (Remember to enter your answer as an ordered pair.)

 

 

Thanks for your help!

 Dec 1, 2019
 #1
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+1

We can use the reflective property of the ellipse.  Since triangle F_1F_1'C is isoceles, we can conctruct the triangle with ray F_1 C.

 

 

We know F_1F'C is isosceles, and EC is a perpendicular bisector, so angle F_1CE = angle F_1'CE (*).

 

We also know

 

Using these properties, we can determine that the other x-intercept is at (9/2,0).

 Dec 1, 2019
 #2
avatar+105989 
+1

 

Thanks guest,

I think I have a different approach.

 

An ellipse has foci at (0,2) and (3,0). The ellipse has two x-intercepts, one of which is the origin. What is the other x-intercept? (Remember to enter your answer as an ordered pair.)

 

The locus of an ellipse is the set of all points such that the sum of the distances to the foci are equal to some constant.

The two loci are     (0,2) and (3,0)       and one point on the ellipse is (0,0)

 

I have always enjoyed solving question using the locus of points that fit a given condition. It is easier becasue you do not need to remember a whole heap of formulas. You only need to know definitions which are usually really simple.

 

The distance from (0,0) to (0,2) is 2

The distance from (0,0) to (3,0) is 3

2+3=5

So the constant is 5.

 

Let (x,y) be a point on the ellipse.

 

\(\text{The distance from (x,y) to (0,2) }=\sqrt{ x^2+(y-2)^2} \\ \text{The distance from (x,y) to (3,0) }= \sqrt{(x-3)^2+y^2} \\so\\ \sqrt{ x^2+(y-2)^2}+ \sqrt{(x-3)^2+y^2}=5\\ \)

Now I need the other x intercept which is where y=0

so sub in y=0

 

\(\sqrt{ x^2+(y-2)^2}+ \sqrt{(x-3)^2+y^2}=5\\ \sqrt{ x^2+(-2)^2}+ \sqrt{(x-3)^2+0^2}=5\\ \sqrt{ x^2+4}+ \sqrt{(x-3)^2}=5\\ \sqrt{ x^2+4}+ |x-3|=5\\ \sqrt{ x^2+4}=5-|x-3|\\ \sqrt{ x^2+4}=5-(x-3)\quad if \;\;x\ge3\quad and \quad \sqrt{ x^2+4}=5+(x-3)\quad if \;\;x<3\\ \sqrt{ x^2+4}=8-x \qquad if \;\;x\ge3\qquad and \qquad \sqrt{ x^2+4}=2+x\quad if \;\;x<3\\ x^2+4=x^2-16x+64 \quad if \;\;x\ge3\quad and \qquad x^2+4=x^2+4x+4\quad if \;\;x<3\\ -60=-16x \quad if \;\;x\ge3\qquad and \qquad 0=4x\quad if \;\;x<3\\ x=\frac{60}{16}=\frac{15}{4}=3.75 \quad if \;\;x\ge3\qquad and \qquad x=0\quad if \;\;x<3\\ so\\ \text{The two x intercepts are 0 and 3.74} \)

 

**   There is a typo in the coding.  The second x intercept it    3.75

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 Dec 1, 2019
edited by Melody  Dec 1, 2019
 #3
avatar+105989 
0

Asker guest,

My answer is different from guests. I have not checked mine super well, there could be an error. 

It is up to you to work out which, if either, of us is correct

 

Feel free to ask questions if you want to.

 Dec 1, 2019
 #4
avatar+105989 
+1

Here is your ellipse

 

 Dec 1, 2019

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