Question

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

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

Answer 1

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# 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# 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`