If we take two identical cones, one upright with its vertex pointing up, another upsidedown with its vertex point down, and touching the vertex of the first one. Take a plane and cut the cones in different way, and we obtain different goemetric figures. If the plane cuts parallel to the base of one of the cone, the resulting plane curve we get is a circle. If the plane cuts parallel to one of the sides of the cones, the resulting plane curve we get is a parabola. If the plane cuts not parallel to the base of the cone, but intersects only one cone, the resulting plane curve is an ellipse. If the plane cuts through both cones, not parallel to any side or base, the resulting plane curve is a hyperbola. Parabola Algebraically, A parabola can be defined as the collection of all points (on a plane) that are equidistant from a given point, called the focus and a given line, called the directrix. Note: Remember that by definition, the distance from a point to a line is the perpendicular distance between the point and the line. Example: Let a > 0 be a real number. Find the equation of the parabola whose focus is (0; a) and whose directrix is the line y = a. By definition, we are looking for all the points (x; y) on the plane whose distance (d1) from the point (focus), (0; a), is the same as its distance (d2) from the line (directrix), y = −a. Using the distance formula, we have: p 2 2 d1 = (x − 0) + (y − a) And looking at the picture and knowing that we want the perpendicular distance, we have: d2 = jy − (−a)j = jy + aj (where a is a positive number) Since we want d1 = d2, we have: p(x − 0)2 + (y − a)2 = jy + aj (x)2 + (y − a)2 = jy + aj2 = (y + a)2 (square both sides) x2 + y2 − 2ay + a2 = y2 + 2ay + a2 (multiply) x2 − 2ay = 2ay (simplify) x2 = 4ay The above example gives us the following: If a > 0, the equation of a parabola with its focus at the point (0; a) and directrix y = −a is given by: 1 4ay = x2 or y = x2 4a Notice this equation contains the origin, (0; 0), and this is the vertex of the parabola. The line that contains the focus and the perpendicular to the directrix, in this case the y−axis, is called the axis of symmetry. The graph of this parabola is a parabola with its vertex at the origin and opens up. Keeping the convention that a > 0, the equation x2 = −4ay gives us a parabola with its directrix being y = a and its focus is the point (0; −a). This is a parabola with vertex at the origin but opens down. The equation y2 = 4ax gives us the equation of the parabola with focus at the point (a; 0) and directrix x = −a. This parabola opens to the right. The equation y2 = −4ax gives us the equation of the parabola with focus at the point (−a; 0) and directrix x = a. This parabola opens to the left. Example: 1 Find the focus, directrix, vertex, and graph the parabola: y = x2 12 Ans: We first change it to the form: 4ay = x2, this gives us 12y = x2 ) 4a = 12 ) a = 3. The focus is at the point (0; 3), the directrix is the line y = −3. The vertex of the parabola is at (0; 0). The graph of the parabola opens up. Example: Find the focus, directrix, vertex, and graph the parabola: −14y = x2. 14 7 Ans: This equation is already in the form we like. We see that a = = , focus 4 2 7 7 is at 0; − , directrix is the line: y = , and vertex is (0; 0). This parabola 2 2 opens down. Example: Find the focus, directrix, vertex, and graph the parabola: 10x = y2 10 5 5 Ans: 4a = 10 ) a = = The focus is on ; 0 , directrix is the vertical line 4 2 2 5 x = − , the vertex is still the origin. This parabola opens to the right. 2 Example: Find the focus, directrix, vertex, and graph the parabola: 1 x = − y2 3 3 Ans: Put it into 4ax = y2 form gives −3x = y2. 4a = 3 ) a = . The focus is at 4 3 3 − ; 0 , directrix is the vertical line x = . Vertex is the origin. This parabola 4 4 opens to the left. Instead of the origin, if the vertex of the parabola is shifted to another point (h; k), with the directrix still parallel to the x− or y−axis, then the form of the equation that would give us the focus and directrix is: (x − h)2 = 4a(y − k) This is the equation of a parabola whose vertex is at the point (h; k), the directrix is the line y = k − a. Focus is at the point (h; k + a). The axis of symmetry is the line x = h. This parabola opens up. (x − h)2 = −4a(y − k) This is the equation of a parabola whose vertex is at the point (h; k). The directrix is the line y = k + a. Focus is the point (h; k − a). The axis of symmetry is the line x = h. This parabola opens down. 4a(x − h) = (y − k)2 This is the equation of a parabola with vertex at (h; k). The directrix is the line x = h − a. Focus is the point (h + a; k). This parabola opens to the right. −4a(x − h) = (y − k)2 This is the equation of a parabola with vertex at (h; k). The directrix is the line x = h + a. Focus is the point (h − a; k). The parabola opens to the left. When we are given a quadratic expression in standard form, we can turn it into this form by using the method of completing the square. Example: Find the vertex, directrix, focus, and graph the parabola: y = 2x2 − 3x + 1 Ans: We complete the square to turn this into the form we like: 1 3 1 y = x2 − x + 2 2 2 1 1 3 y − = x2 − x 2 2 2 1 1 9 3 9 y − + = x2 − x + 2 2 16 2 16 1 1 32 y + = x − 2 16 4 1 1 32 y + = x − 2 8 4 Comparing to the vertex form 4a(y − k) = (x − h)2, we see that the vertex is 3 1 (h; k) = ; − , 4 8 1 1 4a = ) a = 2 8 This is a parabola whose vertex is at 3 1 (h; k) = ; − , 4 8 3 the focus is the point (h; k + a) = ; 0 . 4 Directrix is the line y = k − a, so directrix is the line: 1 1 1 y = − − = − 8 8 4 3 The axis of symmetry is the line x = 4 Example: Find the vertex, focus, directrix, and graph the parabola x = −4y2 − y + 2 Ans: This time we complete the square on y: x = −4y2 − y + 2 1 1 1 − x = y2 + y − 4 4 2 1 1 1 − x + = y2 + y 4 2 4 1 1 1 1 1 − x + + = y2 + y + 4 2 64 4 64 1 33 12 − x + = y + 4 64 8 1 33 12 − x − = y + 4 16 8 Compare this equation with the vertex form: −4a(x − h) = (y − k)2, we see that the vertex (h; k) is 33 1 (h; k) = ; − 16 8 1 1 4a = ) a = 4 16 33 1 1 32 1 1 Focus is the point (h − a; k) = − ; − = ; − = 2; − 16 16 8 16 8 8 Directrix is the line x = h + a. So directrix is the line: 33 1 34 17 x = + = ) x = 16 16 16 8 1 This is a parabola that opens to the left. Axis of symmetry is the line y = − 8 Example: Find the equation of the parabola with vertex (3; 1) and focus (−2; 1) Ans: The vertex and the focus lie on the same line of symmetry, y = 1. The distance between the vertex and the focus is a = j3 − (−2)j = 5. Since the focus is on the left hand side of the vertex, this parabola opens to the left. Its equation is: −4(a)(x − h) = (y − k)2. Since a = 5 and (h; k) = (3; 1), the equation of the parabola is: −4(5)(x − 3) = (y − 1)2 −20(x − 3) = (y − 1)2 The directrix is on the other side of the vertex opposite the focus. The equation of the directrix is x = 8. Example: Find the equation of the parabola with focus (−2; −1) and directrix y = 3. Ans: The axis of symmetry is perpendicular to the directrix. This means the axis of symmetry is parallel to the x−axis, meaning the vertex with have the same x−coordinate as the focus. The vertex of the parabola is halfway between the focus and the directrix, so the vertex is (−2; 1) The distance between the focus and the vertex is a = j1 − (−1)j = 2. This is a parabola that opens up. The general equation is: −4a(y − k) = (x − h)2. Since a = 2 and (h; k) = (−2; 1), the equation for this parabola is: −4(2)(y − 1) = (x − (−2))2 −8(y − 1) = (x + 2)2 Example: Find the equation of the parabola given by the graph. Ans: We see from the graph that the parabola has vertex (1; −3). This is a parabola that opens to the right, so its equation will be of the form: 4a(x − h) = (y − k)2 Knowing the vertex is (1; −3) allows us to write the equation as: 4a(x − 1) = (y − (−3))2 4a(x − 1) = (y + 3)2 5 To find the value of a, we use the additional point, ; 0 from the graph.