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Find the standard form of the conic section $x^2-3x+4xy+y^2+21y-15=0$.

I understand the approach in trying to solve these problems. But the $4xy$ is confusing me. I am not sure of where to start on this one.

UPDATE:

i have found a way of doing this with reference to Canonical form of conic section

Solution:

$x^2-3x+4xy+y^2+21y-15=(x+2y)^2-3y^2-3x+21y$

Let:

$X = x+2y$

$Y=y$

$x = X - 2Y$

Now we substitute:

$X^2-3Y^2-3X+27Y-15=0$

so it is a hyperbola?

Please check.

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2 Answers 2

up vote 1 down vote accepted

Consider the second degree terms: $x^2 + 4xy + y^2$. In general, if the second degree term is $Ax^2 + Bxy + Cy^2$, then the discriminant is $D = B^2 - 4AC$. If the conic section is non-degenerate, then \begin{align} D > 0 & \implies \text{Hyperbola}\\ D = 0 & \implies \text{Parabola}\\ D < 0 & \implies \text{Ellipse} \end{align} Hence, in your case, it can be a hyperbola or a pair of straight lines.

Now factorize the second degree term as follows. $$x^2 + 4xy + y^2 = \underbrace{(x+(2+\sqrt{3}) y)}_{X}\underbrace{(x+(2-\sqrt{3}) y)}_{Y}$$ $$x^2 + 4xy + y^2 -3x + 21 y - 15 = XY + aX + bY - 15 =0$$ Hence, we get that $a+ b= -3$ and $(2+\sqrt{3})a+ (2-\sqrt{3})b= 21$.

This gives us $a = \dfrac32 (-1 + 3 \sqrt{3})$ and $b = \dfrac32(-1 - 3 \sqrt{3})$.

Hence, $$x^2 + 4xy + y^2 -3x + 21 y - 15 = XY + aX + bY - 15 = (X+b)(Y+a) - ab - 15 = 0$$

Hence, $$(X+b)(Y+a) = 15 - \dfrac{117}2 = -\dfrac{87}2$$ in the canonical hyperbola form.

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Thanks man, did it a different way can you check the update please. –  Dreamer78692 Jan 9 '13 at 6:04
    
Looks fine. Similar to this answer. It might be better to write it in canonical form so that it is apparent that it is a hyperbola i.e. in your case $$(X+ \sqrt{3}Y + b)(X - \sqrt{3}Y + a) = -\dfrac{87}2$$ –  user17762 Jan 9 '13 at 6:14

Make substitutions $x=u+h$ and $y=v+k$, viewing $u$ and $v$ as new variables, and viewing $h$ and $k$ as constants. You can solve for the values of $h$ and $k$ that wipe out the $u$- and $v$-terms. (Generally if this is not possible then you will have a parabola, but in your example, this is possible.). This leaves you with $$u^2 + Buv + v^2=C$$ for some numbers $B$ and $C$. This equation is equivalent to the matrix equation$$\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}1&B/2\\B/2&1\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}=C$$ The matrix in the middle is symetric, so it has two orthogonal eigenvectors. Changing variables via rotation to new variables $s$ and $t$ that measure along these eigenvectors yields $$\begin{bmatrix}s&t\end{bmatrix}\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\begin{bmatrix}s\\t\end{bmatrix}=C$$ which is just $$\lambda_1s^2+\lambda_2t^2=C$$ Basic understanding of conic sections will then tell you if you have an ellipse or a hyperbola. Its center is at $(h,k)$ and its axes are parallel to eigenvectors of $\begin{bmatrix}1&B/2\\B/2&1\end{bmatrix}$. If it is an ellipse, its diameters may be determined from $C$, $\lambda_1$, and $\lambda_2$. If it is a hyperbola, the same numbers may be used to determine its asymptotics and the minimum distance between the arms.

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Brilliant... fascinating insight. –  Dreamer78692 Jan 9 '13 at 6:29

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