PARABOLA 1 I NTRODUCTION All along, we have been talking about - PowerPoint PPT Presentation

D AY 136 – E QUATION OF A PARABOLA 1

I NTRODUCTION All along, we have been talking about quadratic equations, graphs of quadratic equations and so on. We would like now to move a notch higher and approach this quadratic equation as a form of application in real life. The graph of the quadratic equation represents a figure called parabola which is used to concentrate rays at a particular point called the focus . This property makes a parabola have a lot of applications in designing appliances for use in industries, hospitals, and even homes.

In this lesson, we are going to derive find the equation of the parabola given the directrix and the focus

V OCABULARY Parabola It is a geometric figure in form on a graph of a quadratic function Focus of a parabola It is a point where all rays that fall on the surface of the parabola meet Line of symmetry of a parabola It is a line that divides the parabola into two equal and overlapping parts

V OCABULARY Vertex It is a point at the intersection of the line of symmetry and the parabola Directrix It is a line perpendicular to the line of symmetry of the parabola far from the vertex as the focus is on the opposite side. It is the generator of the parabola

A parabola It is curved line where all rays illuminated on it are concentrated at a central point called the focus. The parabola is attained when a cone is cut at an angle along its curved surface as shown below. Line along which the cone is cut

Based on how the parabola is cut, any point on it is equidistant from both the focus and the directrix. Line of symmetry focus vertex directrix

Consider the following diagram where the vertex is at the origin with the focus at (0, 𝑞) . Let (𝑦, 𝑧) be any point picked on the parabola. Since the vertex is at the center of the focus and the intersection of the line of symmetry and the directrix, equation of the directrix is 𝑧 = −𝑞. P (𝑦, 𝑧) F (0, 𝑞) V (0,0) directrix D (𝑦, −𝑞) O

𝑦 2 + 𝑧 − 𝑞 2 The distance 𝐺𝑄 = The distance 𝐸𝑄 = 𝑧 + 𝑞 2 Since the two distances above must be equal, we have 𝑦 2 + 𝑧 − 𝑞 2 = 𝑧 + 𝑞 2 𝑦 2 + 𝑧 − 𝑞 2 = 𝑧 + 𝑞 2 𝑦 2 + 𝑧 2 − 2𝑧𝑞 + 𝑞 2 = 𝑧 2 + 2𝑧𝑞 + 𝑞 2 𝑦 2 − 2𝑧𝑞 = +2𝑧𝑞 𝑦 2 = 4𝑞𝑧 If the focus moves to a new point (𝑏, 𝑐) , then the equation would be 𝑦 − 𝑏 2 = 4𝑞(𝑧 − 𝑐) If the parabola is facing in the positive side of the 𝑦 − axis, then the equation would be 𝑧 − 𝑐 2 = 4𝑞(𝑦 − 𝑏)

Example Find the equation of a parabola whose focus in (1,2) and directrix is 𝑧 = −6. Solution The parabola is facing upwards since the focus is above the directrix. The focus are on the same vertical line as vertex, hence its 𝑦 − coordinate is 1. One 𝑧 − coordinate side, the vertex is between the directix and the focus, −6+2 hence its 𝑧 − coordinate is = −2. 2 Thus, 𝑏, 𝑐 = 1, −2 𝑏𝑜𝑒 𝑞 = 2 . Thus, using 𝑦 − 𝑏 2 = 4𝑞(𝑧 − 𝑐) the equation is 𝑦 − 1 2 = 8(𝑧 + 2)

HOMEWORK Find the equation of a parabola whose focus in (0,2) and directrix is 𝑧 = 4.

A NSWERS TO HOMEWORK 𝑦 2 = −4(𝑧 − 3)

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