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I tried completing the square, making it equal to zero, and trying to think of it as a circle, but I still can't prove that these two are equivalent. I tried to factor this in Wolfram Alpha, and it spits out that answer. But I don't know how else to get to it.

MrYouMath
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Whoj Jeong
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  • the quadratic form part $16 x^2 - y^2$ is nondegenerate, Hessian determinant nonzero. This means that the center of the conic section occurs where the gradient is zero, here both $32x = 0$ and $-2y+8 = 0.$ – Will Jagy May 28 '18 at 21:30
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    The point is that it is not a circle. First, it would be a hyperbola. Second, the constant term goes away when you complete the square, so instead of $(4x)^2 - (y+4)^2 = 1$, say, you have $(4x)^2-(y+4)^2 = 0$, which is in fact a union of two lines (the asymptotes of the hyperbolas). Note that, as someone already wrote in an answer, this factors as a difference of squares and gives you what you had in the title. – Ted Shifrin May 28 '18 at 21:34
  • Sorry, could you dumb it down for me? – Whoj Jeong May 28 '18 at 21:35
  • Whoj, look at some of the many questions tagged conic sections, let us see what happens when I put the url here: https://math.stackexchange.com/questions/tagged/conic-sections – Will Jagy May 28 '18 at 21:44
  • @TedShifrin we get a million conic section questions here, i think this student is coming at this from some other direction, I guess contest. The attempt to factor is a bit unusual. – Will Jagy May 28 '18 at 21:46
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    @Will, I went there because the OP said “trying to think of it as a circle.” I think your answer is at way too advanced a level, but ... – Ted Shifrin May 28 '18 at 21:48

3 Answers3

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We have that

$$16x^2-y^2+8y-16 =16x^2-(y^2-8y+16)=(4x)^2-(y-4)^2$$

then recall that

$$A^2-B^2=(A+B)(A-B)$$

user
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  • I already did that but how does (4x-0)(4x+0)-(y-4)(y-4) get me to (4x+y-4)(4x-y+4) – Whoj Jeong May 28 '18 at 21:30
  • @WhojJeong From here $(4x)^2-(y-4)^2$ with $A=4x$ and $B=y-4$ we have $(4x)^2-(y-4)^2=(4x+y-4)(4x-y+4)$. – user May 28 '18 at 21:32
  • Thanks makes so much sense! Thank You! – Whoj Jeong May 28 '18 at 21:36
  • @WhojJeong You are welcome! Since you ar enew here recall that you can evaluate to accept an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user May 28 '18 at 21:43
  • @WhojJeong also, my email is in my profile, if you wish to send me money that would be appreciated – Will Jagy May 28 '18 at 21:48
  • @WillJagy No money are not allowed :P! But accept an answer is an answer among the given is how it works there. Refer also to https://meta.stackexchange.com/q/88535/389429 – user May 28 '18 at 21:52
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    I know. Years ago, I remember a user on Math Overflow saying something similar. His phrasing was interesting (and correct): he said an acceptance was the "coin of the realm" – Will Jagy May 28 '18 at 21:55
  • @WillJagy Many new user here ask for answers and never accept them, I think it is a good habit in these cases give a remainder about that, in particular for new users. – user May 28 '18 at 21:58
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I f you want to be systematic, add in a variable $z$ to make this homogeneous, $$ 16 x^2 - y^2 - 16 z^2 + 8 y z $$ with result (below)
$$ 16 x^2 - y^2 - 16 z^2 + 8 y z = 16 x^2 - (y-4z)^2 = (4x+y-4z)(4x-y+4z) $$

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) $$

==============================================

$$ E_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

==============================================

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) $$

Will Jagy
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$$16x^2-y^2+8y-16=16x^2-(y-4)^2=(4x-(y-4))(4x+(y-4))$$

At the last step I used $a^2-b^2=(a-b)(a+b)$, in which $a= 4x$ and $b=y-4$.

MrYouMath
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