What mistakes were made in solving for the equation of the ellipse?

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The discussion focuses on solving for the standard form of an ellipse given its vertices and a point on the ellipse. The initial attempt at the solution incorrectly handled the equation, particularly in the placement of brackets and factoring. Participants suggest substituting the vertex first to determine the value of a^2, then substituting the point (1,2) to solve for b. The key error identified is in the manipulation of the equation, which led to incorrect results. Properly isolating b after moving constants to one side and cross-multiplying is recommended for a correct solution.
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Homework Statement



an ellipse has the vertices (+&-4, 0)and the point (1,2) lies on the ellipse. find the standard form of the equation of the ellipse .

Homework Equations



(x-h)^2/a^2 + (y-k)/ b^2 = 1

The Attempt at a Solution


(x-0)^2/(4)^2 + (y-0)/b^2 = 1
(1-0)^2/16 + (2-0)^2/b^2 = 1
16b^2 ( 1/16+4/b^2=1)
**b^2 +64/16 = 16b^2
** b^2/16 + 4 = b^2
**b=b/4 + 2
note: *** is where i think i made my mistakes.
so will anyone tell me, what i did wrong and help me with how to finish this problem, thanks .
 
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Substitute the vertex first, then you'll find what a^2 works out to be.
After that, then sub in the point (1,2)
 
louie3006 said:
16b^2 ( 1/16+4/b^2=1)

Here, your brackets are in the wrong place and if you expand you'll see that it doesn't match the equation above it, which means your factoring was off.


\frac{1}{16} + \frac{4}{b^2} = 1

To isolate for b, move all the constants to one side and then cross multiply.
 
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