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avatar+10863 
+1

It will have a vertical axis and the equation looks something like:

 

(x-1)^2/3^2 + (y-2)^2/4^2= 1      I'll have to do a little more review and get back to you....

 

The CENTER of the ellipse is directly between the foci

so center =   1,2    (h,k)

 

Equation for an ellipse looks like:

 

(x-h)^2 / b^2   +  (y-k)^2 / a^2 = 1    where h.k is the center (much like the equation for a circle)

so we have:

 

(x-1)^2 /a^2   + (y-2)^2 /b^2 = 1   we can find b^2 by substituting the point given (4,2)

results in b^2 = 9    or b =3

 

(x-1)^2/3^2 + (y-2)^2/a^2  = 1      Now we need to find a^2

Now....from Purplemath (where I have been reading about ellipses) is this:

The three letters are related by the equation b2 = a2 – c2 or, alternatively (depending on your book or instructor), by the equation b2 + c2 = a2.

(Proving the relationship requires pages and pages of algebraic computations, so just trust me that the equation is true. It can also be shown — painfully — that b is also the length of the semi-minor axis, so the distance across the ellipse in the "shorter" direction is 2b. Yes, the Pythagorean Theorem is involved in proving this stuff. Yes, these are the same letters used in the Pythagorean Theorem. No, this is not the same as the Pythagorean Theorem. Yes, this is very confusing. Accept it, make sure to memorize the relationship before the next test, and move on.)

 

OK   then      b^2 +c^2 = a^2       we know b = 3    c=3 (didstance from center to one of the foci) so we have:

 

3^2 +3^2 = a^2       thus a^2 = 18    and  FINALLY, we have the quation of the ellipse in question:

 

(x-1)^2 / 3^2   +  (y-2)^2 / 18  =1              

 

(there PROBABY is an easier way !!!!   This was my thought process as I was reading through Purplemath learning about ellipses....you might want to read through it too ! )

Here is a desmos graph:

ElectricPavlov 6 hours ago
 #1
avatar+81049 
+2

 

Probably a way easier way to do this with Geometry...but....I didn't see it...so.....

 

By SAS,   triangles BCF  and ABE  are congruent

 

Therefore angle EAB  = angle FBC

And angle DAE   = angle BEA

But angle DAE + angle EAB  = 90

But angle FBC +  angle FBA  = 90

So  angle BEA =  angle FBA

So...by  AA congruence....triangles BEG   and ABG are similar

But angle BGA  = angle EGB...so....each must = 90

 

So triangles BEG an ABG are right triangles

And AB  = 2BE

So  AG  = 2 BG

So AB^2  =  AG^2  + BG^2

AB^2  =  (2BG)^2  + BG^2

AB^2  =  5BG^2

So AB   = sqrt(5)BG    =  BC

And  (1/2)AB   =  [sqrt(5)/2] BG

cos angle ABG  = BG/ AB  =  BG / sqrt(5)BG   =  1/ sqrt(5)

sin angle ABG =2BG/sqrt (5)BG  =  2/sqrt(5)

And angle ABG  = angle CFB

So sin angle ABG  = sin angle CFB

But  angle CFB  and angle GFD are supplementary so their sines are equal

So....sin ABG  = sin GFD     

And GFD is obtuse...so   

So... cos GFD  =  -sqrt [  1  - sin^2(ABG) ]  =  -sqrt [ 1 -  sin^2(GFD) ]  = 

 - sqrt [ 1 - (2/sqrt(5))^2 ]  = -sqrt [ 1 - 4/5]    = -sqrt (1 /5)  =  -1/sqrt (5)

 

And

Triangles ABE  and BCF  are congruent by SAS

EA  = FB     BC  = AB     and BE  = CF =  (1/2)AB  = FD

EA^2  =  BC^2 + BE^2    .... so.....

FB^2  =  AB^2  +  (AB/2)^2

FB^2  =  5BG^2 + AB^2/4

FB^2  =  5BG^2 +  (5/4)BG^2

FB   =   BGsqrt  (5 + 5/4)

FB =  BGsqrt [ 25/4]  =  (5/2)BG

 

And  

FG  =  FB - BG  =     (5/2)BG - BG  =  (3/2)BG

 

So  using the Law of Cosines

 

GD^2  =  FG^2  + FD^2 - 2(FG)(FD)cosGFD

GD^2  =  FG^2  + (AB/2)^2 - 2(FG)(AB/2)cosGFD

GD^2  = (9/4)BG^2 +   (5/4)BG^2 - 2 (3/2)BG * ( sqrt(5)BG/2) * [ - (1/sqrt(5) ) ]

 

GD^2  =  (14/4)BG^2 + (3/2)BG^2

GD^2  =  (7/2)BG^2 +  (3/2) BG^2

GD^2  =  (10/2)BG^2

GD^2 =  5BG^2

GD  =  sqrt(5)BG

DG  =  sqrt(5)BG  =  AB

 

 

cool cool cool

CPhill Jan 21, 2018 7:11:36 AM

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