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Here is the problem :

Eliminate $\varphi$ from the equations

$$a^2y\sin \varphi+b^2x\cos\varphi+ab(a^2\sin^2\varphi+b^2\cos^2\varphi)=0$$

$$ax\sec\varphi-by\csc\varphi=a^2-b^2$$

I am stumped for a plan of attack. The only idea I have is the substitution,

$$\sin\varphi=\frac{2t}{1+t^2}$$ $$\cos\varphi=\frac{1-t^2}{1+t^2}$$

but this leads to two fourth degree equations, and I am unsure how then to proceed. I want to keep the algebra to a manageable form.

The question occurs in Hobson, Treatise on Plane Trigonometry, pg.97 What is the solution intended for this question ?

Update:

the equations reduce to

$$(a^3\sin^2\varphi+b^3\cos^2\varphi)x+a^2(ab^2-a^2+b^2)\sin^2\varphi\cos\varphi+ab^4\cos^3\varphi=0$$ $$(a^3\sin^2\varphi+b^3\cos^2\varphi)y +a^4b\sin^3\varphi+b^2(a^2b+a^2-b^2)\sin\varphi\cos^2\varphi=0$$

OK so where is the miracle ?

Rene Schipperus
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    I am intrigued to know the context of this. The second equation is the standard formula for the equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $(a\cos \varphi,b\sin \varphi)$. Is there some geometrical interpretation of these equations one should be aware of? – David Quinn Oct 22 '21 at 17:57
  • In the version of Hobson's Treatise on archive.org, exercise 45 on page 99 has (after fixing what appears to be an obvious typo) $$a^3y\sin\phi+b^3x\cos\phi+ab(a^2\sin^2\phi+b^2\cos^2\phi)=0 $$ $$a x\sec\phi- b y\operatorname{cosec}\phi=a^2-b^2$$ (Note the cubed $a$ and $b$ in the first equation, which brings dimensional homogeneity in the variables $a$, $b$, $x$, $y$.) Removing $\phi$ here leaves the standard ellipse equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ – Blue Oct 22 '21 at 18:57
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    @Blue I see. I have the second edition. It has the question as asked. So I guess that this was also a mistake and your fixed equation must be the real question. – Rene Schipperus Oct 22 '21 at 19:10
  • @Blue I think your interpretation is likely right, but I cannot get to your final elimination, I get these equations. $$\sec\varphi(a^4\sin^2 \varphi+b^4\cos^2\varphi)\frac{x}{a}-a^4\sin^2\varphi+b^4\cos^2\varphi+2a^2b^2\sin^2\varphi=0$$

    $$\csc\varphi(a^4\sin^2 \varphi+b^4\cos^2\varphi )\frac{y}{b} +a^4\sin^2\varphi-b^4\cos^2\varphi+2a^2b^2\cos^2\varphi=0$$

     – Rene Schipperus Oct 22 '21 at 20:13
  • @ReneSchipperus: Write these as $$\begin{align} (a^2\cos^2\phi+b^2\sin^2\phi)\frac{x}{a}=\cos\phi(\cdots)\ (a^2\cos^2\phi+b^2\sin^2\phi)\frac{y}{b}=\sin\phi(\cdots) \end{align}$$ Square and add. When the dust settles on the right-hand side, what remains will cancel with the coefficient on the left-hand side. – Blue Oct 22 '21 at 20:41
  • @Blue Well yes... I am lost in the dust storm :-) – Rene Schipperus Oct 22 '21 at 20:43
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    @ReneSchipperus: It's ultimately not too terrible. Keep the $\cos\phi$ and $\sin\phi$ separated-out. Squaring the other factors gives (defining $s:=\sin\phi$ and $c:=\cos\phi$) ... $$\begin{align} c^2(a^8s^4+b^8c^4+4a^4b^4c^2+\cdots) \ s^2(a^8s^4+b^8c^4+4a^4b^4s^2+\cdots) \end{align}$$ The first couple of terms from the two lines combine nicely to give $a^8s^4+b^8c^4$. You just have to reduce the rest to $2a^4b^4s^2c^2$, which happens in a pretty straightforward manner. (FYI: I did this by hand myself; just so you know I'm not completely dependent upon Mathematica. :) – Blue Oct 22 '21 at 21:01
  • @Blue Yes I see now it works. One thing bothers me though, should not $\frac{x}{a}=\cos\phi$ and $\frac{y}{b}=\sin \phi$ be a solution to the system, it is a solution to the second but not the first. – Rene Schipperus Oct 22 '21 at 21:02
  • @ReneSchipperus: I don't see any reason to suspect that those should give a solution to the system. Just declare victory and move on. :) ... ("Extended discussion" warning. This will be my last comment.) – Blue Oct 22 '21 at 21:12
  • @Blue Yes you are right. Thanks very much for your help in this problem. I was going to ask if you have any thoughts on David Quinn's comment above. Especially since the eliminant is the equation for an ellipse. What is missing is a geometric interpretation to the first equation. – Rene Schipperus Oct 22 '21 at 21:20

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The algebra will get hairy no matter what. With a computer algebra system, you can attack the two $t$-quartics with the method of resultants or Groebner bases. In Mathematica, I just use the Resultant[] function. The result(ant) is this polynomial: $$\begin{align} 0 &= x^6 b^4 (a^2 - b)^2 \\ &+y^6 a^4 (a - b^2)^2 \\ &+x^4 y^2 a b^2 (a^3 + 2 a^4 b - 6 a^2 b^2 + 2 b^3 + a b^4)\\ &+x^2 y^4 a^2 b (2 a^3 + a^4 b - 6 a^2 b^2 + b^3 + 2 a b^4)\\ &-x^4 a^2 b^4 (a^2 - b) (2 a^2 - 2 b^2 + a^2 b^2 - b^3)\\ &-y^4 a^4 b^2 (a - b^2) (2 a^2 + a^3 - 2 b^2 - a^2 b^2)\\ &-x^2 y^2 a^2 b^2 \left(\begin{array}{c} a^4 + 2 a^5 b - 2 a^2 b^2 + a^4 b^2 + 2 a^5 b^2 - 4 a^3 b^3 \\ - 4 a^4 b^3 + b^4 + a^2 b^4 - 4 a^3 b^4 + 2 a^4 b^4 + 2 a b^5 + 2 a^2 b^5\end{array}\right)\\ &+x^2 a^4b^4 (a^2 - b^2) (a^2 - b^2 + 2 a^2 b^2 - 2 b^3)\\ &+y^2 a^4b^4 (a^2 - b^2) (a^2 + 2 a^3 - b^2 - 2 a^2 b^2)\\ &- a^6 b^6 (a^2 - b^2)^2 \end{align}$$

Blue
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    Yikes. But it is possible that the expression factors, or in some way can be expressed in more compact form. – Rene Schipperus Oct 22 '21 at 16:27
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    @ReneSchipperus: It doesn't factor over the integers. The best hope for these kinds of things is usually "clever grouping", which isn't really an automatic process. I might give it a go in a while. ... I'm a little struck by the lack of dimensional homogeneity: "$,a-b^2,$" and "$,a^2-b,$". yuck ... Are you sure the original equations are correct? – Blue Oct 22 '21 at 16:44
  • Yes the original equations are correct. I have a new approach, solve for $x$ and $y$, then hope something nice happens. The book did this with another question. – Rene Schipperus Oct 22 '21 at 17:00
  • @Blue Is your version of Mathematica fast enough to calculate the deg-9 transformation: Collect[Resultant[x^10 + x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1, x^9+a x^8+b x^7+c x^6+d x^5+e x^4+f x^3+g x^2+m x+n-y, x],y] ? My old version can't, unfortunately. Kindly see this question if interested. – Tito Piezas III Jul 07 '23 at 08:00
  • @TitoPiezasIII: No luck here, I'm afraid. – Blue Jul 07 '23 at 10:21
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    @Blue Ok, thanks. – Tito Piezas III Jul 07 '23 at 10:27