Question 1109030
looks like the equation of an ellipse.


assuming that's what it is, then transform the equation into standard form of an ellipse.


start with 16x^2 + 9y^2 + 64x - 54y + 1 = 0


subtract 1 from both sides to get 16x^2 + 9y^2 + 64x - 54y = -1


group the x's and the y's together to get (16x^2 + 64x) + (9y^2 - 54y) = -1


factor out the coefficients so that the grouped variables start with a coefficient of 1.


you will get 16(x^2 + 4x) + 9(y^2 - 6y) = -1


complete the squares on (x^2 + 4x) to get (x+2)^2 - 4
complete the squares on (y^2 - 6y) to get (y-3)^2 - 9


equation becomes 16 * ((x+2)^2 - 4) + 9 * ((y-3)^2 - 9) = -1


simplify to get 16 * (x+2)^2 - 64 + 9 * (y-3)^2 - 81 = -1


add 64 and 81 to both sides of the equation to get:


16 * (x+2)^2 + 9 * (y-3)^2 = 144


divide both sides of the equation by 144 to get:


16 * (x+2)^2 / 144 + 9 * (y-3)^2 / 144 = 1


this can be written as 16/144 * (x+2)^2 + 9/144 * (y-3)^2 = 1


simplify so that the numerator in the fractions is 1 to get (x+2)^2 / 9 + (y-3)^2 / 16 = 1


this is now in standard form of (x-h)^2 / b^2 + (y-k)^2 / a^2 = 1


the designation of the letter a always goes where the largest denominator is and the designation of the letter b always goes where the smallest denominator is.


that's why the b^2 went under the (x-h)^2 and the a^2 went under the (y-k)^2.


in your equation, you have:


a^2 = 16 which makes a = 4
b^2 = 9 which makes b = 3
center of the ellipse is (h,k) which make the center (-2,3).


the major axis of the ellipse is the axis that is the longest.


a is the distance along the major axis from the center of the ellipse to the vertex of the ellipse.


b is the distance along the minor axis from the center of the ellipse to the co-vertex of the ellipse.


c is the distance along the major axis from the center of the ellipse to the the focus of the ellipse.


c is calculated using the formula c^2 = a^2 - b^2.


in your equation, c^2 is therefore equal to 9 - 4 = 5.


this makes c = sqrt(5).


your important values to the ellipse are now:


a = 4
b = 3
c = sqrt(5)
center = (-2,3)


the graph of your ellipse looks like this:


<img src = "http://theo.x10hosting.com/2018/021002.jpg" alt="$$$" >


here's a reference that should help you understand what's going on.


<a href = "http://www.purplemath.com/modules/ellipse.htm" target = "_blank">http://www.purplemath.com/modules/ellipse.htm</a>


here's my diagram of the main parts of the ellipse that should help you identify what's what on the ellipse.


<img src = "http://theo.x10hosting.com/2018/021003.jpg" alt="$$$" >