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

Question

what is this

Answer 1

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.

Answer 2

The ellipse equation

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.

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.

Compare it to standard form of ellipse

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

b  = length of semi-minor axis = 3.

Standard form of is .