3 Objectives Rotation of Axes General Equation of a Conic The Discriminant
4 Rotation of Axes Earlier we have studied that equation of the form Ax2 + Cy2 + Dx + Ey + F = 0always represents conics.We saw that the graph is always an ellipse, parabola, or hyperbola with horizontal or vertical axes (except in the degenerate cases).
5 Rotation of AxesIn this section we study the most general second-degree equationAx2 + Bxy + Cy2 + Dx + Ey + F = 0We will see that the graph of an equation of this form is also a conic.In fact, by rotating the coordinate axes through an appropriate angle, we can eliminate the term Bxy and then use our knowledge of conic sections to analyze the graph.
6 Rotation of AxesIn Figure 1 the x- and y-axes have been rotated through an acute angle about the origin to produce a new pair of axes, which we call the X- and Y-axes.Figure 1
7 Rotation of AxesA point P that has coordinates (x, y) in the old system has coordinates (X, Y) in the new system.If we let r denote the distance of P from the origin and let be the angle that the segment OP makes with the new X-axis, then we can see from Figure 2 (by considering the two right triangles in the figure) thatX = r cos Y = r sin x = r cos( + ) y = r sin( + )Figure 2
8 Rotation of Axes Using the Addition Formula for Cosine, we see that x = r cos( + )= r (cos cos – sin sin )= (r cos ) cos – (r sin ) sin = X cos – Y sin
9 Rotation of AxesSimilarly, we can apply the Addition Formula for Sine to the expression for y to obtain y = X sin + Y cos .By treating these equations for x and y as a system of linear equations in the variables X and Y, we obtain expressions for X and Y in terms of x and y, as detailed in the following box.
10 Example 1 – Rotation of Axis If the coordinate axes are rotated through 30, find the XY-coordinates of the point with xy-coordinates (2, –4).Solution: Using the Rotation of Axes Formulas with x = 2, y = –4, and = 30, we get
11 Example 1 – Solutioncont’dThe XY-coordinates are (– –1 – ).
12 Example 2 – Rotating a Hyperbola Rotate the coordinate axes through 45° to show that the graph of the equation xy = 2 is a hyperbola.Solution: We use the Rotation of Axes Formulas with = 45° to obtain
13 Example 2 – Solutioncont’dSubstituting these expressions into the original equation gives
14 Example 2 – Solutioncont’dWe recognize this as a hyperbola with vertices (2, 0) in the XY-coordinate system.Its asymptotes are Y = X, which correspond to the coordinate axes in the xy-system (see Figure 3).xy = 2Figure 3
16 General Equation of a Conic The method of Example 2 can be used to transform any equation of the formAx2 + Bxy + Cy2 + Dx + Ey + F = 0into an equation in X and Y that doesn’t contain an XY-term by choosing an appropriate angle of rotation.
17 General Equation of a Conic To find the angle that works, we rotate the axes through an angle and substitute for x and y using the Rotation of Axes Formulas:A(X cos – Y sin )2 + B(X cos – Y sin ) (X sin + Y cos )+ C(X sin + Y cos )2 + D(X cos – Y sin )+ E(X sin + Y cos ) + F = 0
18 General Equation of a Conic If we expand this and collect like terms, we obtain an equation of the formAX2 + BXY + CY2 + DX + EY + F = 0whereA = A cos2 + B sin cos + C sin2 B = 2(C – A) sin cos + B(cos2 – sin2 )
19 General Equation of a Conic C = A sin2 – B sin cos + C cos2D = D cos + E sin E = –D sin + E cos F = F
20 General Equation of a Conic To eliminate the XY-term, we would like to choose so that B = 0, that is,2(C – A) sin cos + B(cos2 – sin2) = 0(C – A) sin 2 + B cos 2 = 0B cos 2 = (A – C) sin 2Double-AngleFormulas for Sineand CosineDivide byB sin 2
21 General Equation of a Conic The preceding calculation proves the following theorem.
22 Example 3 – Eliminating the xy-Term Use a rotation of axes to eliminate the xy-term in the equationIdentify and sketch the curve.Solution: To eliminate the xy-term, we rotate the axes through an angle that satisfies
23 Example 3 – Solutioncont’dThus 2 = 60 and hence = 30. With this value of , we getRotation of Axes Formulas
24 Example 3 – Solutioncont’dSubstituting these values for x and y into the given equation leads toExpanding and collecting like terms, we getDivide by 21
25 Example 3 – Solutioncont’dThis is the equation of an ellipse in the XY-coordinate system.The foci lie on the Y-axis. Because a2 =7 and b2 = 3, the length of the major axis is and the length of the minor axis is
26 Example 3 – Solution The ellipse is sketched in Figure 4. cont’d
27 Example 4 – Graphing a Rotated Conic A conic has the equation64x2 + 96xy + 36y2 – 15x + 20y – 25 = 0(a) Use a rotation of axes to eliminate the xy-term.(b) Identify and sketch the graph.(c) Draw the graph using a graphing calculator.
28 Example 4(a) – SolutionTo eliminate the xy-term, we rotate the axes through an angle that satisfies
29 Example 4(a) – Solution In Figure 5 we sketch a triangle with cont’d
30 Example 4(a) – Solution We see that cont’dWe see thatso, using the Half-Angle Formulas, we get
31 Example 4(a) – Solution The Rotation of Axes Formulas then give and cont’dThe Rotation of Axes Formulas then giveandSubstituting into the given equation, we have
32 Example 4(a) – Solution Expanding and collecting like terms, we get cont’dExpanding and collecting like terms, we get100X2 + 25Y – 25 = 0–4X2 = Y – 1X2 = (Y – 1)SimplifyDivide by 4
33 Example 4(b) – Solutioncont’dWe recognize this as the equation of a parabola that opens along the negative Y-axis and has vertex (0, 1) inXY-coordinates. Since 4p = we have p = so thefocus is and the directrix is Y =Using = cos–1 37° we sketch the graph in Figure 6(a).64x2 + 96xy + 36y2 – 15x + 20y – 25 = 0Figure 6(a)
34 Example 4(c) – Solutioncont’dTo draw the graph using a graphing calculator, we need to solve for y. The given equation is a quadratic equation in y, so we can use the Quadratic Formula to solve for y. Writing the equation in the form36y2 + (96x + 20y) + (64x2 – 15x – 25) = 0we getQuadraticFormula
35 Example 4(c) – Solutioncont’dExpandSimplifySimplify
36 Example 4(c) – Solutioncont’dTo obtain the graph of the parabola, we graph the functionsandas shown in Figure 6(b).64x2 + 96xy + 36y2 – 15x + 20y – 25 = 0Figure 6(b)
38 The DiscriminantIn Examples 3 and 4 we were able to identify the type of conic by rotating the axes. The next theorem gives rules for identifying the type of conic directly from the equation, without rotating axes.
39 The DiscriminantThe discriminant is unchanged by any rotation; for this reason, the discriminant is said to be invariant under rotation.
40 Example 5 – Identifying a Conic by the Discriminant A conic has the equation3x2 + 5xy – 2y2 + x – y + 4 = 0(a) Use the discriminant to identify the conic.(b) Confirm your answer to part (a) by graphing the conic with a graphing calculator.
41 Example 5 – Solution(a) Since A = 3, B = 5, and C = –2, the discriminant isB2 – 4AC = 52 – 4(3)(–2) = 49 > 0So the conic is a hyperbola.(b) Using the Quadratic Formula, we solve for y to get
42 Example 5 – Solutioncont’dWe graph these functions in Figure 7. The graph confirms that this is a hyperbola.Figure 7