Find the equation of ellipse given vertices and focus Check

[aisha]

Find the equation of ellipse given vertices and focus Check please

Hi the question is find the equation of the following ellipse, given vertices at (8,3) and (-4,3) and one focus at (6,3)

Well I drew a digram with the 3 points

First I found the midpoint of the given vertices to get the center of the ellipse I got (h,k) to be (2,3) then I found the distance between the center and the vertex a=6

the only thing I wasnt sure about was how to find b, but this is what I did I found the distance between the two vertices and got 12, I think this is also the legnth of the major axis therefore 2b=12 so b=6

My final equation for the ellipse is

\frac {(x-2)^2} {36} + \frac {(y-3)^2} {36} =1 Help me out is this correct?

 

[dextercioby]

The equation that u've written describes a circle and not an ellipse...

Do'em all again...

Daniel.

 

[Data]

think about your method of finding b and see if you can figure out something wrong with it (hint: using your method, could you ever have b \neq a? Does b really represent the length of the major axis?)

Once you've thought about that, see if you can deduce where the other focus is (remember, ellipses have two of them!)

 

[aisha]

I can probably find where the other focus but how will this help me? How do I find b? Is the rest of the equation correct?

Oh b does not represent the length of the major axis um it represent the length of the minor axis? 2b?


I DONT HAVE A CLUE on HOW TO FIND B! HELPPPPPPP

 

[Data]

The rest is fine. b is the length of the semi-minor axis, so it's given by b^2 = a^2 - c^2 where c is half the distance between the focii.


Do you see why b has this value?

 

[aisha]

The rest is fine. b is the length of the semi-major axis, so it's given by b^2 = a^2 - c^2 where c is half the distance between the focii.


Do you see why b has this value?

See my diagram doesn't have semi-major axis on it that's why I don't know how to find b, um I got the distance between the foci to be 8 half of this is 4 when I plugged a and -c I got
b^2=6^2 =4^2 b = sqrt (20) ? I don't really see why b has this value if this is correct...

 

[Data]

Sorry, as has been corrected now I meant semi-minor (here a is actually the semi-major axis, silly terminology). Your answer for b is correct nonetheless.

Do you remember what at ellipse is?

Given two points P_0 = (x_0, y_0) and P_1 = (x_1, y_1), an ellipse with semi-major axis a and focii P_0 and P_1 is the set of points P = (x,y) such that the sum of the distances from P to P_0 and from P to P_1 is 2a.

Can you sketch this shape? Once you do this, find c using the method I described above, and try to see if you can figure out why b is what it is, by constructing triangles with base c and hypotenuse a within the shape.

 

[aisha]

Ok yes I see the triangle so finally the final equation is

\frac {(x-2)^2} {36} + \frac {(y-3)^2} {20} =1

 

[Data]

Indeed