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9x^2 + 16y^2 - 18x - 64y - 71 = 0 write equation in standard form for the ellipse described

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what is this

asked Dec 6, 2013 in ALGEBRA 2 by angel12 Scholar

2 Answers

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Given equation 9x^2+16y^2-18x-64y-71 = 0

9x^2-18x+16y^2-64y-144+73 = 0

9x^2-18x+16y^2-64y+64+9-144 = 0

9x^2-18x+9+16y^2-64y+64-144 = 0

9(x^2-2x+1)+16(y^2-4y+4)-144 = 0

9(x-1)^2+16(y-2)^2-144 = 0

Divide to each side by 144.

9(x-1)^2/144++16(y-2)^2/144-144/144 = 0/144

(x-1)^2/16+(y-2)^2/9-1 = 0

Add 1 to each sde.

(x-1)^2/16+(y-2)^2/9-1+1 = 0+1

(x-1)^2/4^2+(y-2)^2/3^2 = 1

Compare it to standard form of ellipse (x-h)^2/a^2+(y-k)^2/b^2 = 1

Where a > b.

Standard form to given equation is (x-1)^2/4^2+(y-2)^2/3^2 = 1.

answered Dec 10, 2013 by william Mentor
0 votes

The ellipse equation image

image

image

To change the expressions (x 2- 2x ) and (y 2 - 4y ) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)²to each side of the equation.

image

image

image

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 144 to set it equal to 1.

image

image

image

Compare it to standard form of ellipse image

a 2 > b 2

If the larger denominator is under the "x" term, then the ellipse is horizontal.

center (h, k ) = (1, 2)

a  = length of semi-major axis = 4

= length of semi-minor axis = 3.

Standard form of image is image.

 

answered Jun 5, 2014 by david Expert

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