What are the equations in standard form of the equations #9x^2+16y^2=144# and #25x^2+9y^2-18y-216=0# ?

(In ellipses. I don't know how to convert from general form to standard form.)

1 Answer
marfre
May 6, 2018

#x^2/16 + y^2/9 = 1" "# horizontal ellipse centered at #(0, 0)#

#x^2/9 + (y-1)^2/25 = 1" "# vertical ellipse centered at #(0, 1)#

Explanation:

Given: #9x^2 + 16y^2 = 144# and #25x^2 + 9y^2 -18y -216 = 0#

To point an ellipse or hyperbola in standard form you must complete the square if needed, then divide by the constant on the right side:

Equation 1: #" "(9x^2)/144 + (16y^2)/144 = 144/144#

#(9x^2)/144 + (16y^2)/144 = 1#

Reduce so there isn't a constant in the numerator:

#x^2/16 + y^2/9 = 1" "# horizontal ellipse centered at #(0, 0)#

It's horizontal because the largest denominator is on the #x#-term.

#--------------------#
Equation 2:
Requires completing of the square. First group #x# terms together, #y#-terms together and put the constant on the right:

#(25x^2) + (9y^2 -18y) = 216#

Factor: #25x^2 + 9(y^2 -2y) = 216#

Complete the square on #y# by halving the #-2y# constant #= -1# and adding the 9(-1)^2 on the right that was added to the left when the square was completed:

#25x^2 + 9(y-1)^2 = 216 + 9(-1)^2#

#25x^2 + 9(y-1)^2 = 225#

Divide by the constant on the right side:

#(25x^2)/225 + (9(y-1)^2)/225 = 225/225#

#(25x^2)/225 + (9(y-1)^2)/225 =1#

Reduce so there isn't a constant in the numerator:

#x^2/9 + (y-1)^2/25 =1" "# vertical ellipse centered at #(0, 1)#

It's vertical because the largest denominator is on the #y#-term.

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