### Video Transcript

Consider the conic given by the equation four π₯ squared plus three π¦ squared minus 32π₯ plus six π¦ plus 55 equals zero. Write the equation in standard form. Hence, describe the conic.

First letβs consider what standard form would look like. π₯ minus β squared over π squared plus π¦ minus π squared over π squared equals one. We can also note that under these conditions, point β, π is the center of the conic. And that means we definitely have some rearranging to do. We start by subtracting 55 from both sides of the equation. And then weβll have something that looks like this. Now we want to regroup the terms. The ones with π₯-variables go together. And the ones with π¦-variables go together. Four π₯ squared minus 32π₯ plus three π¦ squared plus six π¦ equals negative 55. Then we notice that both of our π₯ terms have a coefficient thatβs divisible by four.

If we take out the factor of four, weβll be left with four times π₯ squared minus eight π₯. When we look at the π¦-coefficients, we see that theyβre both divisible by three. If we take out the three, weβll have three times π¦ squared plus two π¦. And, again, bring down that negative 55. When we think about our π₯- and π¦-values in standard form, theyβre part of squares. π₯ minus β squared and π¦ minus π squared. To put our equation in this format, we need to do something called completing the square.

When we do that, weβll be able to find four times π₯ plus some value squared plus three times π¦ plus some value squared equals negative 55. These missing values are π over two π, when ππ₯ squared plus ππ₯ plus π equals zero or when ππ¦ squared plus ππ¦ plus π equals zero. We have π₯ squared minus eight π₯. And that means π over two π equals negative eight over two times one, which is π₯ plus negative four. And we can simplify that to say π₯ minus four. Our π¦ portion equals π¦ squared plus two π¦, π equals two and π equals one. Two divided by two times one equals two over two, which is one. By now I hope youβre thinking, but we canβt just add things to one side of the equation. If we add a value to the left side of the equation, we need to add that value to the right side.

But what value exactly did we add to the left side of the equation? You might think, well, we added negative four and one. But thatβs not what we added here. Because this negative four is part of a square. And it must be multiplied by four. To find out what we added to the equation, weβll need to expand both of these squares. For the π₯-values, weβll say four times π₯ minus four times π₯ minus four. Weβll FOIL π₯ times π₯ equals π₯ squared. π₯ times negative four equals negative four π₯. Negative four times π₯ equals negative four π₯. We combine like terms, negative four π₯ minus four π₯ equals negative eight π₯. Remember, weβll need to multiply all of this by four. Four π₯ squared minus 32π₯ plus 64. Four times π₯ minus four squared equals four π₯ squared minus 32π₯ plus 64.

We do the same thing for our π¦ terms. π¦ plus one times π¦ plus one. π¦ squared plus π¦ plus π¦ plus one. We combine the like terms, π¦ plus π¦ equals two π¦. And we multiply all of this by three. Three π¦ squared plus six π¦ plus three. Three times π¦ plus one squared in expanded form is three π¦ squared plus six π¦ plus three. And all of this is equal to negative 55.

Now we go back and look at what we started with. We started with four π₯ squared minus 32π₯ plus three π¦ squared plus six π¦. And that means, to the left side of the equation, weβve added 64 and three. If weβve added 64 and three to the left side, to complete the square for our π₯-variable, we added 64 to the left side. And now weβll add 64 to the right side. To complete the square with our π¦-variable, we added three to the left side. And we need to add three to the right side. Negative 55 plus 64 plus three equals 12.

So we have four times π₯ minus four squared plus three times π¦ plus one squared equals 12. This is much closer to standard form. But in standard form, the equation has to be equal to one. To set this equation equal to one, weβll divide every term by 12. 12 divided by 12 equals one. Four over 12 equals one-third. And we can say π₯ minus four squared over three is the simplified form here. And our π¦ term has the fraction three over 12, which can be reduced to one-fourth. So π¦ plus one squared over four is the simplified form.

This is our equation in standard form. π₯ minus four squared over three plus π¦ plus one squared over four equals one. We noticed that our π₯- and π¦-values have the same sign. This tells us that weβre dealing with an ellipse. We also know that β π is the center of this conic. We have π₯ minus four squared and π¦ plus one squared. Our β would be four. But since weβre dealing with π¦ plus one, our π-value is negative. π equals negative one. The center here is at four, negative one. This is an equation for an ellipse with the center at four, negative one.