How do you graph #4x^2+49y^2+294y+245=0#?

identify the center, vertices, co-vertices, foci, and eccenticity of each

1 Answer
Mar 20, 2018

Answer:

Use the discriminant to identify the equation as an ellipse
Complete the squares to obtain the standard form

Explanation:

In the section entitled [General Cartesian Form](https://en.wikipedia.org/wiki/Conic_section#General_Cartesian_form)# we find the equation:

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

We observe that for the equation #4x^2+49y^2+294y+245=0#:

#A= 4, B = 0, C = 49, D=0, E = 294, and F = 245#

In the section entitled [Discriminant](https://en.wikipedia.org/wiki/Conic_section#Discriminant)# we are told that the:

#"Discriminant" = B^2-4AC#

Substitute #B=0,A=4 and C = 49#

#"Discriminant" = 0^2-4(4)(49)#

#"Discriminant" = -784#

The same section tells us that if the discriminant is negative and #A!=C#, then the conic section is an ellipse, therefore, we must make the equation fit one of two standard forms :

#(x-h)^2/a^2+(y-k)^2/b^2= 1, a>b " [1]"#

#(y-k)^2/a^2+(x-h)^2/b^2= 1, a>b " [2]"#

In either case, we must complete the squares, using the patterns, #(x-h)^2= x^2-2hx+h^2# and #(y-k)^2=y^2-2ky+k^2#.

Subtract 245 from both sides:

#4x^2+49y^2+294y=-245#

The fact that #D=0# tells us that #h = 0#

#4(x-0)^2+49y^2+294y=-245#

Because #C=49#, we multiply the pattern for the y-terms by 49:

#49(y-k)^2=49y^2-98ky+49k^2#

This tells us that we must add #49k^2# to both sides of the equation:

#4(x-0)^2+49y^2+294y+49k^2=-245+49k^2#

We can find the value of k by set the middle term in the right side of the pattern equal to the middle term in the equation:

#-98ky=294y#

#k = -3#

Substitute the left side of the pattern into the left side of the equation and substitute #49k^2 = 441# into the right side of the equation:

#4(x-0)^2+49(y- (-3))^2=-245+441#

Simplify the right side:

#4(x-0)^2+49(y- (-3))^2=196#

Divide both sides of the equation by 196:

#(x-0)^2/49+(y- (-3))^2/4=1#

Write the denominators as squares:

#(x-0)^2/7^2+(y- (-3))^2/2^2=1#

The following is the graph:

graph{(x-0)^2/7^2+(y- (-3))^2/2^2=1 [-7.51, 8.294, -6.807, 1.09]}

The center is the point #(h,k) = (0,-3)#
The vertices are the points #(h-a,k) = (-7,-3)# and #(h+a,k) = (7,-3)#
The covertices are the points #(h,k-b) = (0,-5)# and #(h,k+b) = (0,-1)#
The foci are the points #(h-sqrt(a^2-b^2),k) = (-sqrt53,-3)# and #(h+sqrt(a^2-b^2),k) = (sqrt53,-3)#
The eccentricity is: #sqrt(a^2-b^2)/a = sqrt53/7#