WEBVTT
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Consider the conic given by the equation four π₯ squared plus three π¦ squared minus 32π₯ plus six π¦ plus 55 equals zero.
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Write the equation in standard form.
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Hence, describe the conic.
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First letβs consider what standard form would look like.
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π₯ minus β squared over π squared plus π¦ minus π squared over π squared equals one.
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We can also note that under these conditions, point β, π is the center of the conic.
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And that means we definitely have some rearranging to do.
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We start by subtracting 55 from both sides of the equation.
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And then weβll have something that looks like this.
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Now we want to regroup the terms.
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The ones with π₯-variables go together.
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And the ones with π¦-variables go together.
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Four π₯ squared minus 32π₯ plus three π¦ squared plus six π¦ equals negative 55.
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Then we notice that both of our π₯ terms have a coefficient thatβs divisible by four.
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If we take out the factor of four, weβll be left with four times π₯ squared minus eight π₯.
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When we look at the π¦-coefficients, we see that theyβre both divisible by three.
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If we take out the three, weβll have three times π¦ squared plus two π¦.
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And, again, bring down that negative 55.
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When we think about our π₯- and π¦-values in standard form, theyβre part of squares.
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π₯ minus β squared and π¦ minus π squared.
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To put our equation in this format, we need to do something called completing the square.
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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.
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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.
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We have π₯ squared minus eight π₯.
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And that means π over two π equals negative eight over two times one, which is π₯ plus negative four.
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And we can simplify that to say π₯ minus four.
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Our π¦ portion equals π¦ squared plus two π¦, π equals two and π equals one.
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Two divided by two times one equals two over two, which is one.
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By now I hope youβre thinking, but we canβt just add things to one side of the equation.
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If we add a value to the left side of the equation, we need to add that value to the right side.
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But what value exactly did we add to the left side of the equation?
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You might think, well, we added negative four and one.
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But thatβs not what we added here.
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Because this negative four is part of a square.
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And it must be multiplied by four.
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To find out what we added to the equation, weβll need to expand both of these squares.
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For the π₯-values, weβll say four times π₯ minus four times π₯ minus four.
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Weβll FOIL π₯ times π₯ equals π₯ squared.
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π₯ times negative four equals negative four π₯.
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Negative four times π₯ equals negative four π₯.
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We combine like terms, negative four π₯ minus four π₯ equals negative eight π₯.
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Remember, weβll need to multiply all of this by four.
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Four π₯ squared minus 32π₯ plus 64.
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Four times π₯ minus four squared equals four π₯ squared minus 32π₯ plus 64.
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We do the same thing for our π¦ terms.
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π¦ plus one times π¦ plus one.
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π¦ squared plus π¦ plus π¦ plus one.
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We combine the like terms, π¦ plus π¦ equals two π¦.
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And we multiply all of this by three.
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Three π¦ squared plus six π¦ plus three.
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Three times π¦ plus one squared in expanded form is three π¦ squared plus six π¦ plus three.
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And all of this is equal to negative 55.
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Now we go back and look at what we started with.
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We started with four π₯ squared minus 32π₯ plus three π¦ squared plus six π¦.
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And that means, to the left side of the equation, weβve added 64 and three.
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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.
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And now weβll add 64 to the right side.
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To complete the square with our π¦-variable, we added three to the left side.
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And we need to add three to the right side.
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Negative 55 plus 64 plus three equals 12.
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So we have four times π₯ minus four squared plus three times π¦ plus one squared equals 12.
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This is much closer to standard form.
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But in standard form, the equation has to be equal to one.
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To set this equation equal to one, weβll divide every term by 12.
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12 divided by 12 equals one.
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Four over 12 equals one-third.
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And we can say π₯ minus four squared over three is the simplified form here.
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And our π¦ term has the fraction three over 12, which can be reduced to one-fourth.
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So π¦ plus one squared over four is the simplified form.
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This is our equation in standard form.
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π₯ minus four squared over three plus π¦ plus one squared over four equals one.
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We noticed that our π₯- and π¦-values have the same sign.
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This tells us that weβre dealing with an ellipse.
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We also know that β π is the center of this conic.
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We have π₯ minus four squared and π¦ plus one squared.
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Our β would be four.
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But since weβre dealing with π¦ plus one, our π-value is negative.
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π equals negative one.
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The center here is at four, negative one.
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This is an equation for an ellipse with the center at four, negative one.