MHB Find Equation of Ellipse at (1,2) & (1,8), Minor Axis Length 4

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The equation of the ellipse with center at (1,2), a vertex at (1,8), and a minor axis length of 4 is derived as (x-1)²/16 + (y-2)²/36 = 1. The semi-major axis is determined to be 6, while the semi-minor axis is 2. Participants emphasize the importance of showing work when seeking help to facilitate better guidance. Clarifications are made regarding the lengths of the axes to ensure accurate understanding. The discussion highlights the collaborative nature of problem-solving in mathematics.
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Find equation for center at (1,2) and vertex at (1,8) and Minor axis length of 4?
 
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Hello, schooler!

Did you make a sketch?

Find equation of the ellipse with center (1,2),
and vertex at (1,8) and minor axis length of 4.
Code:
          |
          |   *(1,8)
          |   :
          |   :
          |   :6
          |   :
          |   :   2
      * . | . + . . . *
          |   :(1,2)
    - - - + - : - - - - -
          |   :
          |   :
          |   :
          |   *
          |
We have enough information to write the equation.

. . \frac{(x-1)^2}{4} + \frac{(y-2)^2}{36} \;=\;1
 
schooler said:
Find equation for center at (1,2) and vertex at (1,8) and Minor axis length of 4?

Hello, schooler! :D

We do ask that you show what you have tried so our helpers can see where you are stuck and can help get you unstuck. If you simply post a problem with no work shown, we don't really know how to help, other than perhaps give you hints you have already tried which wastes your time and the time of the helper. Most of our helpers are not going to just work the problem for you, because this does not really get you involved in the process and maximize the "learning moment."
 
soroban said:
\frac{(x-1)^2}{4} + \frac{(y-2)^2}{36} \;=\;1
This should be
\[
\frac{(x-1)^2}{16} + \frac{(y-2)^2}{36}=1
\]
 
Hello, Evgeny!

The semi-major axis has length 6.
Hence: a = 6.

The minor axis has length 4.
The semi-minor axis has has length 2.
Hence: b = 2.

 
Sorry, you are right. I read it as the minor semi-axis has length 4.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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