Question #f97bd

1 Answer
Jan 6, 2018

Answer:

Use the discriminant of the conic section equation to determine what type it is.
Rewrite the equation in its standard form.
Use the attributes of that form to aid in graphing.

Explanation:

The equation

#9x^2+4y^2-18x+16y-1=0#

Is in the General Cartesian form for a conic section:

#Ax^2+Bxy+Cy^2+Dx+Ey+F = 0#

Where #A = 9,B=0,C=4,D=-18,E=16, and F = -1#.

To determine what type of conic section the equation is, one uses the discriminant:

#Delta =B^2-4AC#

  • If #Delta <0# then the equation is that of a circle or an ellipse. If #B=0 and A =C# then it is a circle. Otherwise, it is an ellipse
  • If #Delta = 0# then the equation is that of a parabola.
  • If #Delta > 0# then the equation is that of a hyperbola.

In our case #Delta = 0^2-4(9)(4) < 0# and #A !=C#

Therefore, the equation is that of an ellipse.

There are two standard Cartesian forms for the equation of an ellipse:

  1. #(x-h)^2/a^2+(y-h)^2/b^2 = 1; a>b#
  2. #(y-h)^2/a^2+(x-h)^2/b^2 = 1; a>b#

Equation 1 has a horizontal major axis and the following attributes:

Center: #(h,k)#.
Length of the horizontal axis: 2a
Vertices: #(h-a,k) and (h+a,k)#
Length of the vertical axis: 2b
Co-vertices: #(h,k-b) and (h,k+b)#

Equation 2 has a vertical major axis and the following attributes:

Center: #(h,k)#.
Length of the vertical axis: 2a
Vertices: #(h,k-a) and (h,k+a)#
Length of the horizontal axis: 2b
Co-vertices: #(h-b,k) and (h+b,k)#

To obtain one of the two standard forms, we must complete the square in the equation:

#9x^2+4y^2-18x+16y-1=0#

Move the constant to the right and group the x and y terms together:

#9x^2-18x+4y^2+16y=1#

To make the pattern, #(x-h)^2=x^2-2hx+h^2#, fit the equation, please observe that we must multiply both sides by 9:

#9(x-h)^2=9x^2-18hx+9h^2#

This tells us that we must insert +9h^2 into the x terms on the left and add #9h^2# to the right:

#9x^2-18x+9h^2+4y^2+16y=1+9h^2#

We can set the middle term on the right side of the pattern equal to middle term on the equation, to find the value of h:

#-18hx = -18x#

#h = 1#

We can substitute #9x^2-18x+9(1)^2= 9(x-1)^2# on the left and substitute #9h^2 = 9(1)^2# on the right:

#9(x-1)^2+4y^2+16y=10#

To make the pattern, #(y-k)^2=y^2-2ky+k^2#, fit the equation, please observe that we must multiply both sides by 4:

#4(y-k)^2=4y^2-8ky+4k^2#

This tells us that we must add #4k^2# to both sides of the equation:

#9(x-1)^2+4y^2+16y+4k^2=10+4k^2#

Set the middle term in the right side of the pattern equal to the middle term in the equation and solve for k:

#-8ky = 16y#

#k = -2#

Substitute #4y^2+16y+4(-2)^2 =4(y- (-2))^2# on the left and substitute #4k^2=4(-2)^2# on the right:

#9(x-1)^2+4(y- (-2))^2=26#

Divide both sides by 26:

#9(x-1)^2/26+4(y- (-2))^2/26=1#

Write the denominators as fractions:

#(x-1)^2/(26/9)+(y- (-2))^2/(26/4)=1#

Write the denominators as squares:

#(x-1)^2/(sqrt26/3)^2+(y- (-2))^2/(sqrt26/2)^2=1#

Swap the two terms so that is fits equation 2:

#(y- (-2))^2/(sqrt26/2)^2+(x-1)^2/(sqrt26/3)^2=1#

The above is the equation of an ellipse with a vertical major axis and the following attributes:

Center: #(1,-2)#.
Length of the vertical axis: #sqrt26#
Vertices: #(1,-2-sqrt26/2) and (1,-2+sqrt26/2)#
Length of the horizontal axis: #(2sqrt26)/3#
Co-vertices: #(1-sqrt26/3,-2) and (1+sqrt26/3,-2)#

This should help you to graph the equation.