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Problem:

Given the points $$(0,1),(0,-1),(2,0),(-2,0),(1,1)$$ find the equation for the conic section that passes through these points.

My attempt:

Using the general equation for a conic section, $$ax^2+bxy+cy^2+dx+ey+f=0$$ I inserted the points to make a system of 5 equations, which was reduced to the following, with $a$ as a free variable.

$$b=-a \\c=4a \\d=e=0 \\f=-4a$$

Putting this into the general equation for a conic, I got $$ax^2-axy+4ay^2-4a=0 \ \ (1)$$

My question:

Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of $b^2+4ac$, where I ended up with $b^2-4ac = -15a^2 < 0$ which is supposed to be an ellipse. However, I can't find my way back to the place where I read this. Looking at the points on a graph, it seems visually obvious that this is an ellipse. However, I can't prove it unless the mentioned method is valid.

If it really is an ellipse, how can I algebraically manipulate $(1)$ into the form most known for ellipses?

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  • $\begingroup$ The form most known for ellipses has major and minor axes parallel to the $x$ and $y$ axes, and thus no cross term (i.e. $b = 0$). Without rotating the coordinate system there is no hope of writing this as $\frac{(x-i)^2}{j} + \frac{(y-k)^2}{l} = 1$ $\endgroup$
    – Arthur
    Jan 10, 2015 at 1:52
  • $\begingroup$ @Amzoti - Yes, there is the $b^2-4ac$ bit I was talking about. Using that calculator confirmed what I thought, but I still need to prove that $x^2-xy+4y^2-4=0$ represents an ellipse. Also, any non-zero $a$ will yield the same ellipse. This is obvious since you can factor it out from $(1)$. $\endgroup$
    – Alec
    Jan 10, 2015 at 2:14
  • $\begingroup$ @Arthur - You're right. Drawing the equation in Geogebra shows a slightly rotated ellipse. Could $(1)$ after factoring and striking $a \neq 0$ be the answer to the problem? $\endgroup$
    – Alec
    Jan 10, 2015 at 2:16
  • $\begingroup$ Why not to use the method described in en.wikipedia.org/wiki/… $\endgroup$ Jan 10, 2015 at 4:23
  • $\begingroup$ For finding the equation in the first place, solving is the system requires a guess about which variable to leave "free"; and, of course, there's all that solving work to do. Using the determinant mentioned in this answer, one generates the equation directly ... albeit with the work of expanding a $6\times 6$ determinant. $\endgroup$
    – Blue
    Aug 18, 2018 at 16:04

2 Answers 2

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Any second-degree curve equation can be written as $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0\tag{1}$$ or $$ax^2+2hxy+by^2+2gx+2fy+c=0\tag{2}$$ where $$A,B,C,D,E,E,a,b,c,f,g,h\in\mathbb R$$

To find type of conic and nature of conic we use $\Delta$ and is given by $$\Delta=\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}$$ $$=abc+2fgh-af^2-bg^2-ch^2\tag{3}$$

If $\Delta$ is $0$, it represents a degenerate conic section. Otherwise, it represents a non-degenerate conic section.

Also, the type of conic section that the above equation represents can be found using the discriminant of the equation, which is given by $B^2-4AC$.

Conditions regarding the quadratic discriminant are as follows:

If $\Delta=0$:

$\bullet$ If $h^2-ab\gt0$, the equation represents two distinct real lines.

$\bullet$ If $h^2-ab=0$, the equation represents parallel lines.

$\bullet$ If $h^2-ab\lt0$, the equation represents non-real lines.

If $\Delta\neq0$:

$\bullet$ If $B^2-4AC\gt0$, it represents a hyperbola and a rectangular hyperbola $(A+C=0)$.

$\bullet$ If $B^2-4AC=0$, the equation represents a parabola.

$\bullet$ If $B^2-4AC\lt0$, the equation represents a circle $(A=C,B=0$) or an ellipse $(A\neq C)$. For a real ellipse, $\Big(\frac\Delta{a+b}\lt0\Big)$.

So for the given case, the equation of conic is (after putting all the points in the general equation of conic and finding all the coefficient) $$ax^2-axy+4ay^2-4a=0$$

Comparing above equation with equation (1) and (2) we get the values of required coefficient as $A=a$, $B=-a$ and $C=4a$ so we get the value of $B^2-4AC=-3a^2$ and also from equation (3) we get $\Delta=-15a^2$. Since $\Delta\neq0$ and $B^2-4AC\lt0$ and also $A\neq C$, so as we know, this is the condition of an ellipse.

$\therefore$ Conic $ax^2-axy+4ay^2-4a=0$ is a real ellipse.

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Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of b^2+4ac, where I ended up with b^2−4ac=−15a^2<0 which is supposed to be an ellipse.

Yes, it is b^2-4ac actually.

First, solve the determinant

$\begin{bmatrix}a & b/2 & d/2\\b/2 & c & e/2\\d/2 & e/2 & f\end{bmatrix}$ = p

Now, if:-

  1. p=0 and b^2-4ac>0, it is a pair of distinct straight lines.
  2. p=0 and b^2-4ac=0, it is a either the same line or pair of parrallel lines.
  3. p=0 and b^2-4ac<0, it is a pair of imaginary lines.
  4. p≠0 and b^2-4ac=0, it is a parabola.
  5. p≠0 and b^2-4ac<0, it is an ellipse.
  6. p≠0 and b^2-4ac>0, it is a hyperbola.
  7. p≠0, b=0 and a=c, it is a circle.

Sorry, but I'm not that good with MathJax.

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