# Write equation of translated graph in general form

## Homework Statement

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

## Homework Equations

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

## 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.

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

Mark44
Mentor

## Homework Statement

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

## Homework Equations

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

## 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.

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.

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.

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?

Mark44
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.

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

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?

Mark44
Mentor
The xy term shows up when conic sections (such as an ellipse in this problem) are rotated.
wvcaudill2 said:
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?
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.

HallsofIvy
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.
$$\begin{bmatrix}A & \frac{1}{2}C \\ \frac{1}{2}C & D\end{bmatrix}$$