Problem 33 - Algebra and Trigonometry

Question

Identify each equation without applying a rotation of axes. $$24 x^{2}+16 \sqrt{3} x y+8 y^{2}-x+\sqrt{3} y-8=0$$

Step-by-Step Solution

Answer

The detailed steps will involve identifying parameters A, B, and C from the equation, calculating the discriminant \(B^2 - 4AC\) and then determining whether it's a circle, ellipse, parabola or hyperbola based on the values of the discriminant and comparison of A and C.

Step 1: Identify the form of quadratic equation

The general form of a quadratic equation in two variables x and y is: \(Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0\), where A, B, C, D, E and F are constants. Compare the given equation \(24x^{2} + 16\sqrt{3}xy + 8y^{2} - x + \sqrt{3}y - 8 = 0\) with the general form.

Step 2: Determine the type of conic

For a quadratic equation in two variables, the discriminant \(B^2 - 4AC\) determines the type of conic. If \(B^2 - 4AC > 0\), the equation is that of a hyperbola. If \(B^2 - 4AC = 0\), the equation is that of a parabola. If \(B^2 - 4AC < 0\), then the type of equation depends on whether A equals C or not. If A equals C, it is a circle, else it is an ellipse. Here, A=24, B=16\sqrt{3}, C=8, so calculate \(B^2 - 4AC\).

Step 3: Formulate the Result

If the results equates to zero or a negative number, compare the coefficients A and C to finally determine the type of conic.