Equation of Parabola

 https://www.analyzemath.com/parabola/Equation.html

https://www.analyzemath.com/parabola/Equation.html

https://www.analyzemath.com/parabola/Equation.html

Definition and Equation of a Parabola with Vertical Axis A parabola is the set of all points M(x,y)$M(x,y)$ in a plane such that the distance from M$M$ to a fixed point $$ called the focus is equal to the distance from $$ to a fixed line called the directrix as shown below in the graph. Let us consider a parabola with a vertex $\mathrm{}\mathrm{}$ (the lowest point) at the origin (0,0) as shown in the graph and the focus $\mathrm{}$ on the axis of symmetry (the y axis) with p>0$p>0$. The distance between the points M(x,y)$M(x,y)$ on the parabola and the focus F(0,p)$F(0,p)$ is given by MF=(x−0)2+(y−p)2−−−−−−−−−−−−−−−√$MF=\sqrt{(x-0{)}^2+(y-p{)}^2}$ The distance from point M(x,y)$M(x,y)$ to the directix of equation y=−p$y=-p$ is given by MD=y+p$MD=y+p$ According to the above definition of the parabola these two distances are equal; hence (x−0)2+(y−p)2−−−−−−−−−−−−−−−√=y+p$\sqrt{(x-0{)}^2+(y-p{)}^2}=y+p$ Square both sides and expand the two sides of the equation x2+y2−2py+p2=y2+2py+p2$x^2+y^2-2py+p^2=y^2+2py+p^2$ Group like term 4py=x2$4py=x^2$ Write the equation of the parabola as y$y$ in terms of x$x$. y=14px2$y={\displaystyle \frac{1}{4p}}{x}^2$ Example 1 Point (4,2)$(4,2)$ is on the graph of a parabola with vertex at the origin (0,0)$(0,0)$ and vertical axis. Find the focus of the parabola, graph it and label the focus and graph the directrix. Solution to Example 1 The equation of a parabola with vertical axis at whose vertex is at the origin is given by y=14px2$y={\displaystyle \frac{1}{4p}}{x}^2$ Since (4,2)$(4,2)$ is on the graph of the parabola, the coordinates x=4$x=4$ and y=2$y=2$ satisfy the equation of the parabola. Hence 2=14p(4)2$2={\displaystyle \frac{1}{4p}}(4{)}^2$ Simplify 2=164p$2={\displaystyle \frac{16}{4p}}$ Solve for p$p$ p=2$p=2$ The focus is at the point F(0,2)$F(0,2)$ and the directrix is given by the horizontal line y=−2$y=-2$ as shown in the graph below. We can generalize and write the equation of a parabola at a vertex V(h,k)$V(h,k)$ as follows y=14p(x−h)2+k$y={\displaystyle \frac{1}{4p}}(x-h{)}^2+k$ with vertex V(h,k)$V(h,k)$ and focus F(h,k+p)$F(h,k+p)$ and directrix given by the equation y=k−p$y=k-p$ Example 2 Find the vertex, focus and directrix of the parabola given by the equation y=116x2−14x+94$y={\displaystyle \frac{1}{16}}{x}^2-{\displaystyle \frac{1}{4}}x+{\displaystyle \frac{9}{4}}$ . Solution to Example 2 Rewrite the given equation in standard form by completing the square. factor 1/16$1/16$ out of the terms in x$x$ and x2$x^2$ y=116(x2−4x)+94$y={\displaystyle \frac{1}{16}}(x^2-4x)+{\displaystyle \frac{9}{4}}$ . Complete the square inside the parentheses y=116((x−2)2−22)+94$y={\displaystyle \frac{1}{16}}((x-2{)}^2-2^2)+{\displaystyle \frac{9}{4}}$ Rewrite in standard form y=116((x−2)2−4)−14+94$y={\displaystyle \frac{1}{16}}((x-2{)}^2-4)-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{9}{4}}$ Group like terms y=116(x−2)2+2$y={\displaystyle \frac{1}{16}}(x-2{)}^2+2$ Compare the above equation to the standard form y=14p(x−h)2+k$y={\displaystyle \frac{1}{4p}}(x-h{)}^2+k$ and identify the parameters p$p$, h$h$ and k$k$ 116=14p${\displaystyle \frac{1}{16}}={\displaystyle \frac{1}{4p}}$; solve for p$p$ to obtain p=4$p=4$ h=2$h=2$ and k=2$k=2$ Vertex at V(h,k)=V(2,2)$V(h,k)=V(2,2)$, Focus at F(h,k+p)=F(2,6)$F(h,k+p)=F(2,6)$ , directrix given by y=k−p=−2$y=k-p=-2$ Equation of a Parabola with Horizontal Axis The equation of a parabola with a horizontal axis is written as x=14p(y−k)2+h$x={\displaystyle \frac{1}{4p}}(y-k{)}^2+h$ with vertex V(h,k)$V(h,k)$ and focus F(h+p,k)$F(h+p,k)$ and directrix given by the equation x=h−p$x=h-p$ Example 3 Find the vertex, focus and directrix of the parabola given by the equation x=14y2−y+11$x={\displaystyle \frac{1}{4}}{y}^2-y+11$ . Solution to Example 3 Group the terms in y2$y^2$ and y$y$ and factor 1/4$1/4$ out. x=14(y2−4y)+11$x={\displaystyle \frac{1}{4}}(y^2-4y)+11$ Use the terms y2$y^2$ and y$y$ inside the parentheses and complete the square x=14((y2−2)−22)+11$x={\displaystyle \frac{1}{4}}((y^2-2)-2^2)+11$ Rewrite in standard form y=14((y−2)2)+10$y={\displaystyle \frac{1}{4}}((y-2{)}^2)+10$ Group like terms Compare the above equation to the equation in standard form x=14p(y−k)2+h$x={\displaystyle \frac{1}{4p}}(y-k{)}^2+h$ and identify the parameters p$p$, h$h$ and k$k$ 14p=14${\displaystyle \frac{1}{4p}}={\displaystyle \frac{1}{4}}$ gives p=1$p=1$ h=10$h=10$ and k=2$k=2$ Vertex at V(h,k)=V(10,2)$V(h,k)=V(10,2)$, Focus at F(h+p,k)=F(11,2)$F(h+p,k)=F(11,2)$ , directrix given by x=h−p=9$x=h-p=9$ Interactive Turorial on Equation of a Parabola An app to explore the equation of a parabola and its properties is now presented. The equation used is the standard equation that has the form y=14p(x−h)2+k$y={\displaystyle \frac{1}{4p}}(x-h{)}^2+k$ where h and k are the x- and y-coordinates of the vertex of the parabola and p is a non zero real number. The exploration is carried out by changing the parameters p,h$p,h$ and k$k$ included in the above equation and carry out the activities described below. The default values when you open this page are: p=1,h=2$p=1,h=2$ and k=3$k=3$ Click on the button "Plot Equation" to start. p$p$ =  h$h$ =  k$k$ =  y=14×1(x−2)2+3=0.25(x−2)2+3$$y={\displaystyle \frac{1}{4\times 1}}(x-2{)}^2+3=0.25(x-2{)}^2+3$$ Hover the mousse cursor on the graph or plotted point to read the coordinates. 1 - Start with the default values p=1,h=2$p=1,h=2$ and k=3$k=3$ the button "Plot Equation". Hover the mousse cursor over the graph to trace and read the coordinates of points on the graph, on the focus F or vertex V. a) Use the values of p=1,h=2$p=1,h=2$ and k=3$k=3$ and calculate the coordinates of the focus F$F$, the vertex V$V$ and the equation of the directrix and compare them to the graphical values. b) Select a point M$M$ on the parabola and find the distance MF$MF$ and compare it to the the distance from M$M$ to the directrix.(see definition of parabola above). Are they equal?(or close) 2 - On paper, find the equation of the parabola for the values p=4,h=1$p=4,h=1$ and k=−4$k=-4$. a) Calculate the coordinates of the focus F$F$, the vertex V$V$ and the equation of the directrix b) Calculate the x and y intercepts c) Set the values p=4,h=1$p=4,h=1$ and k=−4$k=-4$ in the app above and then read and check the equation of the parabola, the coordinates of the focus F$F$ and vertex V$V$ and the equation of the directrix. d) Check the x and y intercepts 3 - Exercise: a) On paper, rewrite the equation x2−4x−4y=0$$x^2-4x-4y=0$$ in the form y=14p(x−h)2+k$y={\displaystyle \frac{1}{4p}}(x-h{)}^2+k$ (see example 2 above) b) Identify and find the values of p$p$, h$h$ and k$k$. c) Find the coordinates of the focus F$F$, the vertex V$V$, the x and y intercepts and the equation of the directrix d) Use the app above and check the values found by calculations. If needed, Free graph paper is available. More References and Links to Topics Related to the Equation of the Parabola Tutorial on How Parabolic Dish Antennas work? Tutorial on how to Find The Focus of Parabolic Dish Antennas. Use of parabolic shapes as Parabolic Reflectors and Antennas. Interactive tutorial on how to find the equation of a parabola. Define and Construct a Parabola. Three Points Parabola Calculator. Similar tutorials on circle , Ellipse and the hyperbola can be found in this site.

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