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

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

Mar 20, 2018

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_sectionGeneral_Cartesian_form) we find the equation:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

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

$A = 4 , B = 0 , C = 49 , D = 0 , E = 294 , \mathmr{and} F = 245$

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

$\text{Discriminant} = {B}^{2} - 4 A C$

Substitute $B = 0 , A = 4 \mathmr{and} C = 49$

$\text{Discriminant} = {0}^{2} - 4 \left(4\right) \left(49\right)$

$\text{Discriminant} = - 784$

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

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 , a > b \text{ }$

${\left(y - k\right)}^{2} / {a}^{2} + {\left(x - h\right)}^{2} / {b}^{2} = 1 , a > b \text{ }$

In either case, we must complete the squares, using the patterns, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ and ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$.

Subtract 245 from both sides:

$4 {x}^{2} + 49 {y}^{2} + 294 y = - 245$

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

$4 {\left(x - 0\right)}^{2} + 49 {y}^{2} + 294 y = - 245$

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

$49 {\left(y - k\right)}^{2} = 49 {y}^{2} - 98 k y + 49 {k}^{2}$

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

$4 {\left(x - 0\right)}^{2} + 49 {y}^{2} + 294 y + 49 {k}^{2} = - 245 + 49 {k}^{2}$

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:

$- 98 k y = 294 y$

$k = - 3$

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

$4 {\left(x - 0\right)}^{2} + 49 {\left(y - \left(- 3\right)\right)}^{2} = - 245 + 441$

Simplify the right side:

$4 {\left(x - 0\right)}^{2} + 49 {\left(y - \left(- 3\right)\right)}^{2} = 196$

Divide both sides of the equation by 196:

${\left(x - 0\right)}^{2} / 49 + {\left(y - \left(- 3\right)\right)}^{2} / 4 = 1$

Write the denominators as squares:

${\left(x - 0\right)}^{2} / {7}^{2} + {\left(y - \left(- 3\right)\right)}^{2} / {2}^{2} = 1$

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 $\left(h , k\right) = \left(0 , - 3\right)$
The vertices are the points $\left(h - a , k\right) = \left(- 7 , - 3\right)$ and $\left(h + a , k\right) = \left(7 , - 3\right)$
The covertices are the points $\left(h , k - b\right) = \left(0 , - 5\right)$ and $\left(h , k + b\right) = \left(0 , - 1\right)$
The foci are the points $\left(h - \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(- \sqrt{53} , - 3\right)$ and $\left(h + \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(\sqrt{53} , - 3\right)$
The eccentricity is: $\frac{\sqrt{{a}^{2} - {b}^{2}}}{a} = \frac{\sqrt{53}}{7}$