# Write equation of translated graph in general form

1. Jul 9, 2010

### wvcaudill2

1. The problem statement, all variables and given/known data
Identify the graph of each equation. Write the equation of the translated graph for T(-5,6) in general form. Then draw the graph.

4x2 + 5y2 = 20

2. Relevant equations
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

3. The attempt at a solution
I put the equation instandard form and got :

$$\frac{x^2}{5}$$ + $$\frac{y^2}{4}$$ = 1

This shows that it is a graph of an ellipse with center (0,0).

However, I wasnt sure what to do next. I found the translated points using matrices and adding, but I dont know what to do with this, ot even if I need this information.

2. Jul 9, 2010

### Gear.0

Is it standard form?
let's see,
(standard form) = 0
(some equation) = 1

3. Jul 9, 2010

### Staff: Mentor

If the graph of the ellipse is translated left by 5 units and up 6 units, the center will move from (0, 0) to (-5, 6). How will this affect the equation
$$\frac{x^2}{5} + \frac{y^2}{4} = 1$$?

Tip: write your entire equation in one pair of [ tex] tags.

4. Jul 9, 2010

### wvcaudill2

The equation will then become:

$$\frac{(x+5)^2}{5} + \frac{(y-6)^2}{4} = 1$$

NOw, how can I convert this into general form?

5. Jul 9, 2010

### Staff: Mentor

Multiply both sides by 20, expand (x + 5) and (y - 6)2, and bring everything to one side, with zero on the other side.

6. Jul 9, 2010

### wvcaudill2

Ok, I get the following:

$$4x^2 + 5y^2 + 40x - 60y +260 = 0$$

This looks good, except there is no "Bxy" term. When would this term appear?

Looking through my older notes, I found a few problems where i converted from standard form to general form before. However, it looks like that way I did it was by setting up a system of 3 equations, and somehow solving for the variables D, E, and F. Can this method I mentioned by used on anything other than circles?

Lastly, will the method you showed me work with all types of conic sections?

7. Jul 9, 2010

### Staff: Mentor

The xy term shows up when conic sections (such as an ellipse in this problem) are rotated.
I don't understand what you're saying here. To convert from general form to standard form (if I remember these terms correctly), you complete the square in the x terms and in the y terms. If you're talking about the general form with an xy term, it's been a very long time since I've done that, so can't commit on the method you describe.

8. Jul 10, 2010

### HallsofIvy

Staff Emeritus
If you have a conic section with $Ax^2+ Bxy+ Cy^2$ (and other non-squared terms) with C not 0, you can get rid of it in either of two ways:

1) Let $x= x' cos(\theta)- y' sin(\theta)$, $y= x' sin(\theta)+ y' cos(\theta)$, replace x and y in your formula by that, and solve for a value of $\theta$ that make Bxy= 0. $\theta$ is the angle through which the axes are rotated and the entire equation with x', y' instead of x, y will have no "xy" term.

2) Find the eigenvalues and eigenvectors of the matrix
$$\begin{bmatrix}A & \frac{1}{2}C \\ \frac{1}{2}C & D\end{bmatrix}$$

The eigenvectors will point along line y= ax and y= bx, that are parallel to the axes of the conic and substituting x'= ax- y, y'= bx- y will get rid of the "xy" term.