SOLUTION: Discuss the graph of the conic section below: 4x^2 +25y^2 -8x+150y+129=0

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Question 1090812: Discuss the graph of the conic section below:
4x^2 +25y^2 -8x+150y+129=0
Found 2 solutions by ikleyn, natolino_2017:
Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
This conic section is an ellipse.

To identify its elements  (axes, semi-axes, eccentricity, focuses) you need to get its standard equation.

For it, complete the squares separately for x-terms and y-terms of the given equation.

To be familiar with the subject and on how to do it, read the lessons in this site
    - Ellipse definition, canonical equation, characteristic points and elements

    - Standard equation of an ellipse
    - Identify elements of an ellipse given by its standard equation
    - Find the standard equation of an ellipse given by its elements

    - General equation of an ellipse
    - Transform a general equation of an ellipse to the standard form by completing the square (*)
    - Identify elements of an ellipse given by its general equation

The major source of this list is the lesson marked (*).


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Conic sections: Ellipses. Definition, major elements and properties. Solved problems".



Answer by natolino_2017(77)   (Show Source): You can put this solution on YOUR website!
4x^2 + 25y^2 -8x+150y + 129 = 0
4(x^2-2x) +25(y^2+6y) + 129 = 0
4((x-1)^2 - 1) +25((y+3)^2 -9) +129 =0
4(x-1)^2 +25(y+3)^2 = -129 + 4 + 225 = 100
(x-1)^2/5^2 + (y+3)^2/2^2 = 1
Ellipse with center on (1,-3) , a=5, b=2 so c=sqrt(25-4)=sqrt(21).
So Focus are F1= (1+sqrt(21),-3) and F2=(1-sqrt(21),-3)
@natolino_
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