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

[tex] \frac {(x-2)^2} {36} + \frac {(y-3)^2} {36} =1 [/tex] 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 [itex]b[/itex] and see if you can figure out something wrong with it (hint: using your method, could you ever have [itex] b \neq a[/itex]? Does [itex]b[/itex] 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. [itex]b[/itex] is the length of the semi-minor axis, so it's given by [itex]b^2 = a^2 - c^2[/itex] where [itex]c[/itex] is half the distance between the focii.


Do you see why [itex]b[/itex] has this value?

 

[aisha]

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


Do you see why [itex]b[/itex] 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
[tex] b^2=6^2 =4^2 [/tex] 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 [itex]a[/itex] is actually the semi-major axis, silly terminology). Your answer for [itex]b[/itex] is correct nonetheless.

Do you remember what at ellipse is?

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

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

 

[aisha]

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

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

 

[Data]

Indeed