What conic section is represented by the equation 3x^2+2y^2-24x+30=0? Explain how you know?

Answer 1

To identify the conic section represented by the equation $3x^2+2y^2-24x+30=0$, we can follow the steps to rearrange and analyze it.

First, we can rearrange the equation into a more standard form. Let's group the terms related to $x$ and $y$:

$$3x^2-24x+2y^2+30=0$$

Next, we can isolate the terms involving $x$ and $y$:

$$3x^2-24x+2y^2=-30$$

Now, we will complete the square for the $x$ terms. The $x$ terms are $3x^2-24x$. To complete the square, factor out the coefficient of $x^2$ (which is 3):

$$3(x^2-8x)$$

Now, complete the square inside the parentheses. Take half of $-8$ (which is $-4$), square it (getting 16), and then add and subtract it inside the parentheses:

$$3(x^2-8x+16-16)=3((x-4{)}^2-16)=3(x-4{)}^2-48$$

Substituting this back into the equation gives us:

$$3(x-4{)}^2-48+2y^2=-30$$

Now, move $-48$ to the right-hand side:

$$3(x-4{)}^2+2y^2=-30+48$$

This simplifies to:

$$3(x-4{)}^2+2y^2=18$$

Now divide the entire equation by 18 to put it in standard form:

$$\frac{3(x-4{)}^2}{18}+\frac{2y^2}{18}=1$$

Simplifying this gives:

$$\frac{(x-4{)}^2}{6}+\frac{y^2}{9}=1$$

This equation is now in the standard form of an ellipse:

$$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$$

where $(h,k)$ is the center of the ellipse, and $a^2$ and $b^2$ are the squares of the semi-axis lengths.

Since both terms are positive and the equation resembles that of an ellipse, we conclude that the conic section represented by the original equation $3x^2+2y^2-24x+30=0$ is an ellipse.