For which $k$ does $kx^2+(k+2)y^2-2x-4y-1=0$ represent a parabola, hyperbola, or ellipse?

Question

Here is a parametric conic. They want me to discuss for which values of $k$ this is a parabola, an hyperbola, or an ellipse:

$$kx^2 +(k+2)y^2 - 2x - 4y -1 = 0$$

I wanted to approach this by using the discriminant delta, hence:

$$\delta = \begin{pmatrix} k & 0 & -2\\ 0 & k+2 & -4\\ -2 & -4 & -1 \end{pmatrix}$$

Setting $\delta$ to zero should be giving us for which values the conics degenerate. After this we classify whether the conic is an ellipse, a parabola or an hyperbola by setting the discriminant $\gamma$:

$$\gamma = \begin{pmatrix} k & 0\\ 0 & k+2 \end{pmatrix}$$

greater than, less than, or equal to zero

Is this discussion enough to find for which values of $k$ the parametric conic is an hyperbola, parabola or ellipse?

I'm asking because the solutions of this problem are completely different from the solutions found by calculating discriminants:

• for $-2 < k < 0$, it's an hyperbola,
• for $k = -2$ and $k = 0$, it's a parabola,
• for $-(7+ \sqrt{41})/2 \le k < -2$ and $k > 0$, it's an ellipse

Especially in the case of the ellipse, I don't know where those numbers are coming from, what am I doing wrong?

Answer 1

We can do this the "easy" (not that easy) way or the "hard" (not too hard) way ...

I: Since the conic section has its axes parallel to the coordinate axes ( $ B \ xy = 0 $ ) , we can investigate the coefficients of its equation after "completing the squares". For $ \ kx^2 - 2x + (k+2)y^2 - 4y \ = \ 1 \ \ , $ we have

$$ k· \left(x^2 \ - \ \frac{2}{k} \ x \ + \ \frac{1}{k^2} \right) \ + \ (k+2) · \left(y^2 \ - \ \frac{4}{k+2} \ y \ + \ \frac{4}{(k+2)^2} \right) \ = \ 1 \ + \ \frac{1}{k} \ + \ \frac{4}{k+2} $$

$$ \Rightarrow \ \ k· \left(x \ - \ \frac{1}{k} \right)^2 \ + \ (k+2) · \left(y \ - \ \frac{2}{k+2} \right)^2 \ = \ \ \frac{k^2 \ + \ 7k \ + \ 2}{k·(k+2)} $$

$$ \Rightarrow \ \ \frac{\left(x \ - \ \frac{1}{k} \right)^2}{\left[\frac{k^2 \ + \ 7k \ + \ 2}{k^2 \ · \ (k+2)} \right] } \ \ + \ \ \frac{\left(y \ - \ \frac{2}{k+2} \right)^2}{\left[\frac{k^2 \ + \ 7k \ + \ 2}{k \ · \ (k+2)^2} \right] } \ = \ \ 1 \ \ , $$

putting the equation into "standard form" [note well the difference between the two denominators in the ratios for each term].

The denominator ratios have four "special points". The values $ \ k = 0 \ $ and $ \ k = -2 \ $ cause them to be "undefined". The quadratic polynomial $ \ k^2 + 7k + 2 \ = \ \left( k + \frac72 \right)^2 \ - \ \frac{41}{4} \ $ has the two zeroes $ \ \kappa_{+} \ = \ \frac12·(-7 + \sqrt{41}) \ \approx \ -0.298 \ \ $ and $ \ \kappa_{-} \ = \ \frac12·(-7 - \sqrt{41}) \ \approx \ -6.702 \ \ . $ The real-number line is thus divided into the intervals $ \ k < \kappa_{-} \ \ , \ \ \kappa_{-} < k < -2 \ \ , \ \ -2 < k < \kappa_{+} \ \ , \ \ \kappa_{+} < k < 0 \ \ \text{and} \ \ k > 0 \ \ . $ We then have five cases to examine.

$ \mathbf{ k > 0 : } \quad $ All factors in both denominators are positive, so both terms are positive, giving us the equation of an ellipse;

$ \mathbf{ \kappa_{+} < k < 0 : } \quad $ $ \ k^2 + 7k + 2 \ > \ 0 \ $ , but the denominator of the second term has become negative, making the second term negative, so this is the equation of a "vertical" hyperbola;

$ \mathbf{ -2 < k < \kappa_{+} : } \quad $ now $ \ k^2 + 7k + 2 \ < \ 0 \ $ and the denominator of the second term is negative, so the first term becomes negative and the second term positive, yielding the equation of a "horizontal" hyperbola;

$ \mathbf{ \kappa_{-} < k < -2 : } \quad $ $ \ k^2 + 7k + 2 \ < \ 0 \ $ still, so both denominators become positive again, producing the equation of an ellipse;

$ \mathbf{ k < \kappa_{-} : } \quad $ $ \ k^2 + 7k + 2 \ > \ 0 \ $ again, causing both denominators to be negative, which means both terms are negative and hence cannot sum to $ \ 1 \ $ -- this equation has no real solutions.

For $ \ k = 0 \ \ \text{and} \ k = -2 \ \ , $ the conic equation reduces to $ \ 2y^2 - 4y \ = \ 2x + 1 \ \ $ or $ \ -2x^2 - 2x \ = \ 4y + 1 \ \ , $ which are the equations of "horizontal" and "vertical" parabolas, respectively. We have degenerate conics for $ \ k = \ \kappa_{+} \ \ \text{and} \ \ k = \kappa_{-} \ \ , $ about which we'll say more with the "determinant" method. One other item of note is that the conic can never be a circle, since $ \ k \neq k+2 \ \ . $

$$ \ \ $$

II: The method of matrix representation for conic sections is presented here, so I will simply refer to it and apply the relevant details. We specify (as you have done) two matrices

$$ \mathsf{A_{33}} \ = \ \left[ \begin{array}{cc} A & B/2 \\ B/2 & C \end{array} \right] \ = \ \left[ \begin{array}{cc} k & 0 \\ 0 & k + 2 \end{array} \right] \ \ \text{and} $$ $$ \mathsf{A_Q} \ = \ \left[ \begin{array}{ccc} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{array} \right] \ = \ \left[ \begin{array}{ccc} k & 0 & -1 \\ 0 & k+2 & -2 \\ -1 & -2 & -1 \end{array} \right] \ \ . $$

We want to work (mostly) with the determinants of these matrices, which are $ \ \det \mathsf{A_{33}} \ = \ k·(k+2) \ \ $ and $ \ \det \mathsf{A_Q} \ = \ k·(-k-2-4) + (-1)·(k+2) \ = \ -(k^2 + 7k + 2) \ \ . $ The "special points" and five intervals discussed above are again of interest here.

We have degenerate conics for $ \ \det \mathsf{A_Q} \ = \ 0 \ : $

for $ \ k \ = \ \kappa_{+} \ , \ \det \mathsf{A_{33}} < 0 \ \ , $ which corresponds to a pair of lines of opposite slopes (hyperbola "degenerated" to asymptotes), while for $ \ k \ = \ \kappa_{-} \ , \ \det \mathsf{A_{33}} > 0 \ \ , $ indicating a single point (ellipse "degenerated" to its center).

We'll then look at the non-degenerate cases ( $ \det \mathsf{A_Q} \neq 0 \ ) \ : $

$ \mathbf{ \det \mathsf{A_{33}} < 0 \ : } \quad $ the conic is a hyperbola, but it switches from "vertical" $ ( \ \kappa_{+} < k < 0 \ \Rightarrow \ \det \mathsf{A_Q} < 0 \ $ ) to "horizontal" ( $ \ -2 < k < \kappa_{+} \ \Rightarrow \ \det \mathsf{A_Q} > 0 \ $ ) ;

$ \mathbf{ \det \mathsf{A_{33}} = 0 \ : } \quad $ the conic is a parabola ( $ \ k = 0 \ \ , \ \ k = -2 \ $ ) ; and

$ \mathbf{ \det \mathsf{A_{33}} > 0 \ : } \quad $ the conic is a ellipse -- with $ \ A + C \ = \ k + (k+2) \ = \ 2·(k+1) \ \ , $ it is a real ellipse for $ \ k > 0 \ \Rightarrow \ (A + C) · \det \mathsf{A_Q} < 0 \ \ $ and for $ \ \kappa_{-} < k < -2 \ \Rightarrow \ (A + C) · \det \mathsf{A_Q} < 0 \ \ , $ becoming an imaginary ellipse for $ \ k < \kappa_{-} \ \Rightarrow \ (A + C) · \det \mathsf{A_Q} > 0 \ \ . $

Answer 2

The second invariant sign $I_2 = ab-h^2 = k(k-2)=f(k) $ which alone determines the conic type is plotted on y-axis .The other constants of classical conic equation do not matter.

$$ k=(0,-2).. parabolas; (-2<k<0).. hyperbolas; (k>0, k<-2)...ellipses;$$