![]() Answerįind x x and y, y, where x = x ′ cos θ − y ′ sin θ x = x ′ cos θ − y ′ sin θ and y = x ′ sin θ + y ′ cos θ. X = x ′ cos θ − y ′ sin θ and y = x ′ sin θ + y ′ cos θ x = x ′ cos θ − y ′ sin θ and y = x ′ sin θ + y ′ cos θįinding a New Representation of an Equation after Rotating through a Given Angleįind a new representation of the equation 2 x 2 − x y + 2 y 2 − 30 = 0 2 x 2 − x y + 2 y 2 − 30 = 0 after rotating through an angle of θ = 45°. īecause u = x ′ i ′ + y ′ j ′, u = x ′ i ′ + y ′ j ′, we have representations of x x and y y in terms of the new coordinate system. u = ( x ' cos θ − y ' sin θ ) i + ( x ' sin θ + y ' cos θ ) j Factor by grouping. u = i x ' cos θ − i y ' sin θ + j x ' sin θ + j y ' cos θ Apply commutative property. u = i x ' cos θ + j x ' sin θ − i y ' sin θ + j y ' cos θ Distribute. u = x ′ i ′ + y ′ j ′ u = x ′ ( i cos θ + j sin θ ) + y ′ ( − i sin θ + j cos θ ) Substitute. U = x ′ i ′ + y ′ j ′ u = x ′ ( i cos θ + j sin θ ) + y ′ ( − i sin θ + j cos θ ) Substitute. It may be represented in terms of its coordinate axes. Because A C > 0 A C > 0 and A ≠ C, A ≠ C, the graph of this equation is an ellipse.įigure 5 Relationship between the old and new coordinate planes.Ĭonsider a vector u uin the new coordinate plane. Because A = C, A = C, the graph of this equation is a circle.Ī = −25 A = −25 and C = −4. We can determine that the equation is a parabola, since A A is zero.Ī = 3 A = 3 and C = 3. The graph of this equation is a hyperbola.Ī = 0 A = 0 and C = 9. Identify the graph of each of the following nondegenerate conic sections. Identifying a Conic from Its General Form On the other hand, the equation, A x 2 + B y 2 + 1 = 0, A x 2 + B y 2 + 1 = 0, when A and B are positive does not represent a graph at all, since there are no real ordered pairs which satisfy it. The degenerate case of a hyperbola is two intersecting straight lines: A x 2 + B y 2 = 0, A x 2 + B y 2 = 0, when A and B have opposite signs. For example, the degenerate case of a circle or an ellipse is a point:Ī x 2 + B y 2 = 0, A x 2 + B y 2 = 0, when A and B have the same sign. That is because the equation may not represent a conic section at all, depending on the values of A, B, C, D, E, and F. Notice the phrase “may be” in the definitions. If B does not equal 0, as shown below, the conic section is rotated. If B = 0, the conic section will have a vertical and/or horizontal axes. If either A A or C C is zero, then the graph may be a parabola.If A A and C C are nonzero and have opposite signs, then the graph may be a hyperbola.If A A and C C are equal and nonzero and have the same sign, then the graph may be a circle.If A A and C C are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse.Identify the values of A A and C C from the general form.A x 2 + B x y + C y 2 + D x + E y + F = 0. ![]() Rewrite the equation in the general form, A x 2 + B x y + C y 2 + D x + E y + F = 0.Given the equation of a conic, identify the type of conic. So from 0 degrees you take (x, y) and make them negative (-x, -y) and then you've made a 180 degree rotation.\) When you rotate by 180 degrees, you take your original x and y, and make them negative. If you have a point on (2, 1) and rotate it by 180 degrees, it will end up at (-2, -1) We do the same thing, except X becomes a negative instead of Y. If you understand everything so far, then rotating by -90 degrees should be no issue for you. ![]() Our point is as (-2, -1) so when we rotate it 90 degrees, it will be at (1, -2)Īnother 90 degrees will bring us back where we started. What about 90 degrees again? Same thing! But remember that a negative and a negative gives a positive so when we swap X and Y, and make Y negative, Y actually becomes positive. Our point is at (-1, 2) so when we rotate it 90 degrees, it will be at (-2, -1) ![]() What if we rotate another 90 degrees? Same thing. So from 0 degrees you take (x, y), swap them, and make y negative (-y, x) and then you have made a 90 degree rotation. When you rotate by 90 degrees, you take your original X and Y, swap them, and make Y negative. If you have a point on (2, 1) and rotate it by 90 degrees, it will end up at (-1, 2) In case the algebraic method can help you:
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