Math 25Fall 2004Rotation of AxesIn sections 10.2-4, we found that every equation of the form(1) Ax2+ Cy2+ Dx + Ey + F = 0,with A and C not both 0, can be transformed by completing the square into a standardequation o f a translated conic section. Thus, the graph of this equation is either a parabola,ellipse, or hyperbola with axes parallel to the x and y-axes (there is also the possibilitythat there is no graph or the gra ph is a “degenerate” conic: a point, a line, or a pair oflines). In this section, we will discuss the equation of a conic section which is rotated byan angle θ, so the axes are no longer paral lel to the x and y-axes.To study rotated conics, we first look at the rela tionship b etween the axes of the conicand the x and y-axes. If we label the axes o f the conic by u and v, then we can describethe conic in the usual way in the uv-coo rdinat e system with an equation of the form(2) au2+ cv2+ du + ev + f = 0Label the axis in the first q uadrant by u, so that the u-axis is obtained by rotating thex-axis through an angl e θ, wi th 0 < θ < π/2 (see Figure 1). Likewise, the v-axis i sobtained by rotati ng the y-axis through the angle θ. Now if P is any point with usualpolar coordinates (r, α), then the polar coordinates of P in the uv system are (r, α − θ)(again, see Figure 1). Thus,u = r cos(α − θ) = r(cos α cos θ + sin α sin θ)= r(cos α) cos θ + (r sin α) sin θ = x cos θ + y sin θand(3)v = r sin(α − θ) = r(sin α cos θ − cos α sin θ)= r(sin α) cos θ − (r cos α) sin θ = y cos θ − x sin θThese two equations in (3) describe how to find the uv-coordinates of a point from itsxy-coordinat es.rα-θαθPxyuvFigure 1Using a similar derivation, if we start with the uv-coordinates, then the xy-coordinatesare given by the equations(4) x = u cos θ − v sin θ and y = v cos θ + u sin θNow if we substitute the expressions for u and v from (3) into equation (2), we will endup with an equation of the form(5) Ax2+ Bxy + Cy2+ Dx + Ey + F = 0In other words, a “mi xed” xy term has been introduced.Example 1. Starting with t he ellipse 4u2+9v2= 36 in the uv-system, we obtain equation(5) in the xy-system as follows:4(x cos θ + y sin θ)2+ 9(y cos θ − x sin θ)2= 36⇒ 4(x2cos2θ + 2xy cos θ sin θ + y2sin2θ) + 9(y2cos2θ − 2xy cos θ sin θ + x2sin2θ) = 36⇒ (4 cos2θ + 9 sin2θ)x2− (10 cos θ sin θ)xy + (4 sin2θ + 9 cos2θ)y2= 36If we know that the uv-system is obtained from the xy-system by a rotation of θ =π3radians, for example, then cos θ =12and sin θ =√32, and the final equation is then314x2−5√32xy +214y2= 36.Conversely, we can show that every equation of the form (5) represents a conic section.To do this, we show t hat this equation is really just the equation of a rotated conic. Inother words, we want to apply the conversion formulas (4) for a suitable angle θ so thatthe new uv equat ion has the form (2).First, we can assume that B 6= 0, for otherwise there is nothing to do. Now substituteformula s (4) into equation (5):A(u cos θ − v sin θ)2+B(u cos θ − v sin θ)(v cos θ + u sin θ) + C(v cos θ + u sin θ)2+D(u cos θ − v sin θ) + E(v cos θ + u sin θ) + F = 0Expanding this expression and collecting like t erms yields the equationau2+ buv + cv2+ du + ev + f = 0witha = A cos2θ + B sin θ cos θ + C sin2θb = −2A sin θ cos θ + B(cos2θ − sin2θ) + 2C sin θ cos θc = A sin2θ − B sin θ cos θ + C cos2θd = D cos θ + E sin θe = −D sin θ + E cos θf = FOur goal is for b to be equal to 0. Using the double angle formulassin 2θ = 2 sin θ cos θ and cos 2θ = cos2θ − sin2θwe obtainb = B cos 2θ + (C − A) sin 2θand it follows that b = 0 if(6) cot 2θ =A − CBSince B 6= 0, this equation can always be solved, and θ can be chosen so that 0 < θ < π/2.Example 2. Consider the equation x2+ 2xy + y2+ 4√2x − 4√2y = 0. A = 1, C = 1,and B = 2, so formula (6) impli es that cot 2θ = 0. Therefore cos 2θ = 0, so 2θ = π/2(remember that 0 < θ < π/2, so 0 < 2θ < π) and t hus θ = π/4. It then follows from (4)that x = u/√2 − v/√2 and y = v/√2 + u/√2. Substituting these values into the givenequation and then simplifying the result eventually yields 2u2− 8v = 0, or u2= 4v. Thisis a parabola with vertex at the origin, axis u = 0 (i.e., t he v-axis), focus (u, v) = (0, 1),and directrix v = −1. Now using formula s (4), it follows that the xy-coordinates of thefocus are (x, y) = (−1/√2, 1/√2), the axis is y = −x (since θ = π/4, the v-axi s is theline through the origin with slope equal to tan(3π/4) = −1), and the directrix is theline y = x −√2 (the line contains the point (u, v) = (0, −1), which has xy-coordinates(x, y) = (1 /√2, −1/√2), a nd the slope is 1 since it is parallel to the u-axis (which hasslope tan θ = tan(π/4) = 1)). See Figure 2.Figure 2Example 3. Consider the equation 5x2+ 4√3xy + 9y2+ 3√3x −3y = 30. A = 5, C = 9,and B = 4√3, so formula (6) implies t hat cot 2θ = −1/√3. Therefore tan 2θ = −√3, so2θ = 2π/3 (remember that 0 < θ < π/2, so 0 < 2θ < π) and thus θ = π/3. It t hen followsfrom (4) that x = u/2 −√3v/2 and y = v/2 +√3u/2. Substituting these values into thegiven eq uat ion and t hen simplifying the result eventually yields 11u2+ 3v2− 6v = 30.Completing the square, we obtain 11u2+ 3(v − 1)2= 33. This is an ellipse w ith center(u, v) = (0, 1), major axis on the line u = 0, and minor ax is on the line v = 1. The lengthof the major axi s is 2√11 and the length of the minor axis is 2√3. Thus, the vertices are(u, v) = (0, 1±√11), and the foci are (u, v) = (0, 1±√8). To figure out the xy-coordinates,first note (using formulas (4)) that the center has xy-coordinates (x, y) = (−√3/2, 1 …
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