SOLUTION: How do I graph an ellipse with the equation of (((x-3)^2)/9)+(((y-4)^2)/4)=1 ?

Question 545027: How do I graph an ellipse with the equation of (((x-3)^2)/9)+(((y-4)^2)/4)=1 ?
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
1_ Find the center
2_ Draw the axes
3_ Sketch the ellipse around the axes
4_ If expected, calculate c, and plot the foci on the graph.
is given in a very convenient form, clearly showing the center and axes, which are parallel to the x- and y-axes.
If the major axis is parallel to the x-axis, or to the y-axis, an ellipse is easy to recognize as such, and has no term in xy. This is the case with your ellipse. Easy (or else I would not be tackling it).
Your textbook will tell you to look for h, k, a, and b, and then calculate c. That kind of alphabet soup works well to solve problems if you have the book open in front of you. It does not help your understanding much, though.
You can look at the problem differently. An ellipse is sort of a stretched circle, that has been stretched longer in the direction of the major axis.
<--> would mean a circle centered at (3,4), with radius 3. Its points would be at distance=3 from that center in all directions.
<--> would mean a circle centered at (3,4), with radius 2. Its points would be at distance=2 from that center in all directions.
is the equation of an ellipse centered at (3,4).
The axes are the vertical line <-->, and the horizontal line <-->, which of course, cross at the center.
Where does the ellipse intercepts those axes?
When <-->,
<-->, so .
That gives you points at a distance of 2 above and below the center: (3,2) and (3,6).
When <-->,
<-->, so .
That gives you the vertices of the ellipse, points at a distance of 3 to the left and right of the center: (0,4) and (6,4).

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