What Are the Key Steps to Solve These Ellipse Problems?

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To solve the ellipse problems, the first step is to convert the equations into standard form. For the first problem, the equation 9x^2 + 16y^2 = 144 simplifies to (x^2/16) + (y^2/9) = 1, revealing the center at the origin and the lengths of the major and minor axes. In the second problem, the equation 4x^2 + (y+1)^2 = 9 indicates the center is at (0, -1). Understanding the standard form of an ellipse is crucial for identifying key characteristics. Including textbook references in future posts may enhance clarity and context for problem-solving.
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Problem I:

(The coeffiecients throw me off, I don't know what I'm supposed to do with them)
9x^2 + 16y^2 = 144

Determine:
a) coodinates of the centre
b) lengths of the major and minor axes




Problem II:

Sketch a graph of the ellipse

4x^2 + (y+1)^2 = 9




PS: For future posts would it be of any help to include the textbook referred to etc?

In this case:

Mathpower 12: WE
Problem I - p. 150 #9
Problem II - p. 150 #19

PPS: I've also italicized any comments about my attemts - since people seem to like to help out people making an attempt.
 
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trigger352 said:
Problem I:

(The coeffiecients throw me off, I don't know what I'm supposed to do with them)
9x^2 + 16y^2 = 144

Determine:
a) coodinates of the centre
b) lengths of the major and minor axes

You have to get it in standard form first. Divide both sides by 144.

That gives:

\frac{x^2}{16} + \frac{y^2}{9} = 1

a = length of the major axis
b= length of the minor axis
 
The clue in both the exercises is to recognize the form of an ellipse with its centre in the origin:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

As mentioned a and b are the lengths of the axes.


In the second exercise the centre is not in origo but (x=0,y=-1).
 
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