11/11/04 bh 113 Page3 PARABOLAS Parabola Vertex (0, 0) Concept Equation Example Parabola with vertex (0, 0) and vertical axis x2 = 4py p > 0: opens upward p < 0: opens downward Focus: (0, p) Directrix: y = - p x2 = - 2y has 4p = - 2 or p = - The parabola opens downward with Practice. The equation of a standard parabola is y2 =4ax, where a is an arbitrary constant. P [r. 1, θ. Parametric co-ordinates of Parabola For a parabola, the equation is y 2 = -4ax. Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y 2 = 16x. Parametric curves in the plane 1. This indicates how strong in your memory this concept is. Preview; Assign Practice; Preview. Progress % Practice Now . The idea of parametric equations. Computer algebra systems can also plot in parametric mode. SECTION 10.1 Conics and Calculus 695 Parabolas A parabola is the set of all points that are equidistant from a fixed line called the directrix and a fixed point called the focus not on the line. Parametric equations of the parabola x 2 = 4ay with the vertex A at the origin and the focus F(0, a), and of its translation (x-x 0) 2 = 4a(y-y 0) with the vertex A(x 0, y 0) and the focus F(y 0, y 0 + a) written Plane Curves and Parametric Equations 717 corresponds to −1 ≤ y ≤ 2, so that the plane curve is the portion of the parabola indicated in Figure 9.1, where we have also indicated a number of points on the curve. Note that \(t = \frac{3}{4}\) is the value of \(t\) that give the vertex of the parabola and is not an obvious value of \(t\) to use! Parabola Notes for IIT JEE, Download PDF! In general, a point . 0] do satisfy the equation: r. 2 = (−2) 2 = 4, 4 cos θ = 4 cos 0 = 4. Solution The center is halfway at (3,7). EXAMPLE 1 Find the circle that has a diameter from (1,7) to (5, 7). The parametric equation of the parabola is x = t2 + 1, y = 2t + 1. Section 9.3 The Parabola 903 Solution The given equation,is in the standard form so We can find both the focus and the directrix by finding Divide both sides by 4. PARABOLA LEVEL-I *1. Note: The other forms of the parabola with latus rectum 4a... 3. Because is positive, the parabola, with its symmetry, opens to the right.The focus is 3 units to the right of the vertex, (0, 0). For To better organize out content, we have unpublished this concept. 1. In fact, this is a good example of why just using values of \(t\) to sketch the graph is such a bad way of getting the sketch of a parametric curve. • Each value of the parameter, when evaluated in the parametric equations, corresponds to a point along the curve of the relation. y 2 = 4ax. Note in Figure 10.3 that a parabola is symmetric In general, a point . 1 1.7 (a) to (d) The latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola (Fig. CONIC SECTIONS 189 Standard equations of parabola The four possible forms of parabola are shown below in Fig. The equation of its directrix is (A) x = 0 (B) x + 1 = 0 (C) y = 0 (D) none of these *2. equation has linear terms -2hx and -2ky-they disappear when the center is (0,O). ARC LENGTH, PARAMETRIC CURVES 59 Answer: The given points correspond to the values t = 1 and t = 2 of the parameter, so: L = Z 2 1 sµ dx dt ¶2 µ dy dt ¶2 dt = Z 2 1 p (2t)2 +(3t2)2 dt Z 2 1 √ 4t2 +9t4 dt Z 2 1 t √ 4+9t2 dt 1 18 Z 40 13 √ udu (u = 4+9t2) 1 27 £ 40 3/2−13 1 27 (80 √ 10−13 13). Changing Parametric Equations to Cartesian Equations Parabolas (x = 2at, y = at2): x = 2at t = x 2a (1) y = at2 Sub in (1): y = 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ a x a = a x 4 2 ⇒ x2 =4ay, which is the equation of the parabola, vertex (0, 0), focus (0, a). Based on the equations, it is probably not surprising that the graph is a parabola. Create Assignment. Comments 1.
Finding Parametric Equations of Parabolas % Progress . Standard Equation of Parabola. 1 1.7). Students compare the standard equations and then predict how the general equation will look if it is representing a parabola. Assign to Class.

This page will be removed in future. The midpoint between the focus and the directrix is the vertex, and the line passing through the focus and the vertex is the axis of the parabola. You can graph parametric equations on your TI graphing calculator by switching to parametric mode.

y x 20 40 t = 0 t = 1 t = 1.5 t = 2 20 40 60 FIGURE 9.2 Path of projectile More All Modalities; Share with Classes. As long as you know the coordinates for the vertex of the parabola and at least one other point along the line, finding the equation of a parabola is as simple as doing a little basic algebra.