Conic Complex

Geometry Level 4

f ( x , y ) = n x 2 + ( n m 2 ) x y + m y 2 + 4 x + 2 y + 1 = 0 g ( x , y ) = n x 2 + ( n + m ) x y + m y 2 + 4 x + 2 y + 1 = 0 h ( x , y ) = n x 2 + ( n − m ) x y + m y 2 + 4 x + 2 y + 1 = 0 \begin{aligned} f(x , y) &= nx^2 + \left(\dfrac{nm}{2}\right)xy + my^2 + 4x + 2y + 1 = 0\\ g(x , y) &= nx^2 + (n + m)xy + my^2 + 4x + 2y + 1 = 0\\ h(x , y) &= nx^2 + (n - m)xy + my^2 + 4x + 2y + 1 = 0 \end{aligned}

Let the graphs f , g , h f, g, h be a parabola, a hyperbola, and an ellipse, respectively.

If n n and m m are positive integers satisfying the constraints above, compute 2 n + m 2n + m .


The answer is 12.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

Relevant wiki: Discriminant of a Conic Section

In a form of A x 2 + B x y + C y 2 + D x + E y + F = 0 Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 , we can determine the nature of the graph by computing its quadratic discriminant: B 2 − 4 A C B^2 - 4AC .

For parabola, B 2 − 4 A C = 0 B^2 - 4AC = 0 , so ( n m 2 ) 2 − 4 n m = 0 (\dfrac{nm}{2})^2 - 4nm = 0 . Thus, n m = 16 nm = 16 .

As a result, there are 5 possible pairs of ( n , m ) (n , m) : ( 1 , 16 ) , ( 2 , 8 ) , ( 4 , 4 ) , ( 8 , 2 ) , ( 16 , 1 ) (1 , 16), (2 , 8), (4 , 4), (8 , 2), (16 , 1) .

Then for hyperbola, B 2 − 4 A C > 0 B^2 - 4AC > 0 . However, for n = m = 4 n = m = 4 , B 2 − 4 A C = ( 4 + 4 ) 2 − 4 ( 4 ⋅ 4 ) = 0 B^2 - 4AC = (4 + 4)^2 - 4(4\cdot 4) = 0 , which is not applicable.

Finally, for ellipse, B 2 − 4 A C < 0 B^2 - 4AC < 0 . However, for ( n , m ) = ( 1 , 16 ) (n , m) = (1 , 16) or ( 16 , 1 ) (16 , 1) , B 2 − 4 A C = ( 16 − 1 ) 2 − 4 ( 16 ⋅ 1 ) = 225 − 16 = 209 > 0 B^2 - 4AC = (16 - 1)^2 - 4(16\cdot 1) = 225 - 16 = 209 > 0 , which is also not applicable.

Therefore, only 2 scenarios are left: ( n , m ) = ( 2 , 8 ) (n , m) = (2 , 8) or ( n , m ) = ( 8 , 2 ) (n , m) = (8 , 2) .

However, when we rewrite the first equation, the new coefficients will be as shown:

a x 2 + 2 h x y + b y 2 + 2 g x + 2 f y + c = 0 ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0

In case of n = 8 n = 8 and m = 2 m = 2 , the equation is:

f ( x , y ) = 8 x 2 + 2 ( 4 ) x y + 2 y 2 + 2 ( 2 ) x + 2 ( 1 ) y + 1 = 0 f(x , y) = 8x^2 + 2(4)xy + 2y^2 + 2(2)x + 2(1)y + 1 = 0

Then we can evaluate the discriminant of the equation as:

Δ = ∣ a h g h b f g f c ∣ = a b c + 2 f g h − a f 2 − b g 2 − c h 2 , \Delta =\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}= abc+2fgh-a{ f }^{ 2 }-b{ g }^{ 2 }-c{ h }^{ 2 },

Thus, Δ = ∣ 8 4 2 4 2 1 2 1 1 ∣ = 16 + 16 − 8 − 8 − 16 = 0 , \Delta =\begin{vmatrix} 8 & 4 & 2 \\ 4 & 2 & 1 \\ 2 & 1 & 1 \end{vmatrix}= 16+16-8-8-16 = 0, .

For a parabola and hyperbola, Δ ≠ 0 \Delta \neq 0 while for a real ellipse, Δ a + b < 0 \dfrac{\Delta}{a+b} < 0 . Therefore, n = 2 n = 2 and m = 8 m = 8 .

Checking the discriminants for the other graphs, we will get:

f ( x , y ) = 2 x 2 + 2 ( 4 ) x y + 8 y 2 + 2 ( 2 ) x + 2 ( 1 ) y + 1 = 0 f(x , y) = 2x^2 + 2(4)xy + 8y^2 + 2(2)x + 2(1)y + 1 = 0 ; Δ = ∣ 2 4 2 4 8 1 2 1 1 ∣ = − 18 \Delta =\begin{vmatrix} 2 & 4 & 2 \\ 4 & 8 & 1 \\ 2 & 1 & 1 \end{vmatrix}= -18 .

g ( x , y ) = 2 x 2 + 2 ( 5 ) x y + 8 y 2 + 2 ( 2 ) x + 2 ( 1 ) y + 1 = 0 g(x , y) = 2x^2 + 2(5)xy + 8y^2 + 2(2)x + 2(1)y + 1 = 0 ; Δ = ∣ 2 5 2 5 8 1 2 1 1 ∣ = − 23 \Delta =\begin{vmatrix} 2 & 5 & 2 \\ 5 & 8 & 1 \\ 2 & 1 & 1 \end{vmatrix}= -23 .

h ( x , y ) = 2 x 2 + 2 ( − 3 ) x y + 8 y 2 + 2 ( 2 ) x + 2 ( 1 ) y + 1 = 0 h(x , y) = 2x^2 + 2(-3)xy + 8y^2 + 2(2)x + 2(1)y + 1 = 0 ; Δ = ∣ 2 − 3 2 − 3 8 1 2 1 1 ∣ = − 39 \Delta =\begin{vmatrix} 2 & -3 & 2 \\ -3 & 8 & 1 \\ 2 & 1 & 1 \end{vmatrix}= -39 .

All values are checked according to all graphs. Hence, 2 n + m = 2 × 2 + 8 = 12 2n + m = 2\times 2 + 8 = \boxed{12} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Remember to check that the cubic discriminant is non-zero, before applying B 2 − 4 A C B^2 - 4AC .

Nicely done! (+1) I did not do it so systematically; I just completed some squares.

I believe there is a small gap in your solution and also in the link you provide. To get a (real) ellipse, you want Δ < 0 \Delta<0 , not merely Δ ≠ 0 \Delta \neq 0 . Luckily, Δ = − 39 \Delta=-39 for the function h h , so, we are safe.

Otto Bretscher - 5 years, 1 month ago

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Thank you for your feedback, sir. Honestly I just learned the discriminant of conic sections from the wiki page, but again we are safe. ;)

Worranat Pakornrat - 5 years, 1 month ago

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Maybe you can write something like "We want Δ ≠ 0 \Delta \neq 0 for a parabola or hyperbola, and Δ < 0 \Delta<0 for an ellipse."

Otto Bretscher - 5 years, 1 month ago

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@Otto Bretscher – Thank you for suggestion, sir. If I'm not mistaken, for a real ellipse, the negative discriminant is determined as Δ a + b \dfrac{\Delta}{a+b} where a , b a, b are coefficients of squared x , y x, y . Is that correct?

Worranat Pakornrat - 5 years, 1 month ago

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@Worranat Pakornrat – a = n a=n and b = m b=m are assumed to be positve, but it is still worth stating the condition in full generality.

Otto Bretscher - 5 years, 1 month ago

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