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

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The equation of the ellipse with a center at (1,2), a vertex at (1,8), and a minor axis length of 4 is given by the formula \(\frac{(x-1)^2}{16} + \frac{(y-2)^2}{36} = 1\). The semi-major axis length is 6, denoted as \(a = 6\), while the semi-minor axis length is 2, denoted as \(b = 2\). This equation is derived from the standard form of an ellipse centered at a point with specified axis lengths.

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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.
 

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