Problem 13
Question
Determine the equation of the given conic in \(X Y\)-coordinates when the coordinate axes are rotated through the indicated angle. $$x^{2}+2 \sqrt{3} x y-y^{2}=4, \quad \phi=30^{\circ}$$
Step-by-Step Solution
Verified Answer
The equation in rotated coordinates is \(X^2 - Y^2 = \frac{16}{5}\).
1Step 1: Identify Rotation Angle and Coefficients
The given angle of rotation is \(\phi = 30^{\circ}\). The given equation of the conic is \(x^2 + 2\sqrt{3}xy - y^2 = 4\). We need to identify the coefficients: \(A = 1\), \(B = 2\sqrt{3}\), and \(C = -1\). These coefficients will be used in our transformation formulae.
2Step 2: Calculate Rotation Formula Components
The formulas for transforming coordinates involve trigonometric functions of \(\phi\). We have:- \(\cos(\phi) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}\)- \(\sin(\phi) = \sin(30^{\circ}) = \frac{1}{2}\). Now we can use these values during transformation.
3Step 3: Calculate New Coefficients in Rotated System
Using the trigonometric values found in step 2, calculate the new coefficients using:\[A' = A\cos^2(\phi) + B\cos(\phi)\sin(\phi) + C\sin^2(\phi),\]\[B' = 2(A - C)\cos(\phi)\sin(\phi) + B(\cos^2(\phi) - \sin^2(\phi)),\]\[C' = A\sin^2(\phi) - B\cos(\phi)\sin(\phi) + C\cos^2(\phi).\]Substituting in the values:\[A' = 1\left(\frac{3}{4}\right) + 2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - 1\left(\frac{1}{4}\right) = \frac{5}{4}\]\[B' = 2(1)\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) + 2\sqrt{3}\left(\frac{3}{4} - \frac{1}{4}\right) = 0\]\[C' = 1\left(\frac{1}{4}\right) - 2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - 1\left(\frac{3}{4}\right) = -\frac{5}{4}\].
4Step 4: Write the Equation in New Coordinates
Using the values calculated in Step 3, the equation of the conic in the new \(XY\)-coordinate system is of the form:\[A'X^2 + B'XY + C'Y^2 = D,\] where \(D = 4\). Since \(B' = 0\), substitute to get:\[\frac{5}{4}X^2 - \frac{5}{4}Y^2 = 4.\] Dividing through by \(\frac{5}{4}\), the equation simplifies to:\[X^2 - Y^2 = \frac{16}{5}.\]
5Step 5: Conclusion
The new equation, after rotation by \(30^{\circ}\), represents an unchanged form but in transformed coordinates. It maintains its structure as \(X^2 - Y^2 = \frac{16}{5}\), signaling a hyperbola.
Key Concepts
Coordinate RotationTrigonometric IdentitiesHyperbolic Equations
Coordinate Rotation
When we need to analyze conic sections, such as ellipses, parabolas, or hyperbolas, sometimes the shape appears skewed because of an oblique orientation. To simplify these equations and gain insights easily, we rotate the coordinate system. Coordinate rotation involves transforming the original axes to new, rotated ones. This transformation is especially useful to eliminate the cross-product term,
- When performing coordinate rotation, we aim to align the new X and Y axes in such a way that any linear terms in the product of variables, such as the XY term, are removed or simplified.
- The rotation angle, \(\phi\), indicates by how much and in which direction (clockwise or counterclockwise) the rotation should occur.
Trigonometric Identities
Trigonometric identities play a crucial role in simplifying calculations during coordinate rotation. They allow us to use known angles in formulas to transform coordinates effectively:
- One fundamental identity is \(\cos^2(\phi) + \sin^2(\phi) = 1\), reflecting the Pythagorean theorem.
- For a specific example like \(\phi = 30^\circ\), it helps to know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\) and \(\sin(30^\circ) = \frac{1}{2}\).
- These values are used as components in the transformation equations, affecting the coefficients of our new, rotated conic equation.
Hyperbolic Equations
Not all conic sections are ellipses or circles; some are hyperbolas. A hyperbolic equation often presents itself with a characteristic form, such as \(X^2 - Y^2 = K\), where \(K\) is a constant.
- In hyperbolic equations, one variable is subtracted from another rather than both being added, which is typical for ellipses or circles.
- This structure signals the presence of asymptotes, creating the "open ends" of a hyperbola.
- When rotating to new coordinates, if an equation retains this structure (as in the exercise solved), it still represents a hyperbola, now simply oriented differently in the plane.
Other exercises in this chapter
Problem 12
Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$x^{2}=y$$
View solution Problem 13
Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$x^{2}-y^{2}=1$$
View solution Problem 13
Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$x^{2}+4 y^{2}=16$$
View solution Problem 13
Find the center, foci, vertices, and asymptotes of the hyperbola. Then sketch the graph. $$\frac{(x+1)^{2}}{9}-\frac{(y-3)^{2}}{16}=1$$
View solution