Cfrgtkky

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?

Any second-degree curve equation can be written as

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0tag{1}$$
or
$$ax^2+2hxy+by^2+2gx+2fy+c=0tag{2}$$
where $$A,B,C,D,E,E,a,b,c,f,g,hin mathbb R$$


To find type of conic and nature of conic we use $Delta$ and is given by

$$Delta=
left|
begin{array}
&a&h&g\
h&b&f\
g&f&c\
end{array}
right|
=abc+2fgh-af^2-bg^2-ch^2tag{3}
$$


If $Delta$ is Zero, 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 $h^2-ab gt 0$, the equation represents two distinct real lines.

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

$bullet$ If $h^2-ab lt 0$, the equation represents non-real lines.



If $Delta neq 0$:


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

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

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



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


Comparing above equation with equation (1) & (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 $Deltaneq 0$ and $B^2-4AC lt 0$ and also $Aneq C$, so as we know this is the condition of an Ellipse.



Therefore conic $ax^2-axy+4ay^2-4a=0$ is a Real Ellipse (For Real Ellipse the condition is $frac Delta {a+b} lt 0$).

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

$begin{bmatrix}a & b/2 & d/2\b/2 & c & e/2\d/2 & e/2 & fend{bmatrix}$ = 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.