RS Aggarwal Solutions for Class 11 Maths Chapter 22 Parabola
R S Aggarwal Solutions for Class 11 Maths Chapter 22. Parabola. Now. Focus : F(a
9.3 The Parabola
Write as a quadratic equation in and then use the quadratic formula to express in terms of Graph the resulting two equations using a graphing utility in a.
Conic Sections
of the equations of a line. In this Chapter we shall study about some other curves
Introduction to conic sections
Standard equation for non-degenerate conic section circle x2 + y2 = a2 ellipse x2 a2 + y2 b2 = 1 parabola y2 - 4ax = 0 hyperbola x2.
CONIC SECTIONS
18-Apr-2018 The equation of a circle with radius r having ... Let the equation of the parabola be y2 = 4ax and P(x y) be a point on it. Then the.
Finding the Focal Point
www.ies.co.jp/math/java/conics/focus/focus.html You can use the following equation to determine the ... The formula for a parabola is f = x /4a.
The Parabola and the Circle - San Antonio
The primary focal chord formula is
The Arc Length of a Parabola Let us calculate the length of the
Grinshpan. The Arc Length of a Parabola. Let us calculate the length of the parabolic arc y = x2 0 ? x ? a. According to the arc length formula
conic sections
of the equations of a line. In this Chapter we shall study about some other curves
Exploring Space Through Math
This problem applies mathematical principles in NASA's human spaceflight. use time as a parameter in parametric equations. ... the next parabola.
General Equation of a Parabola - University of Minnesota
Standard Equation of a Parabola k= A(x h)2andx h= A(y k)2 Form of the parabola x2 = y opens upwardx2 = y opens downwardy2 = x opens to the righty2 = x opens to the left Vertex at (h;k) Stretched by a factor of Avertically fory=x2andhorizontally forx=y2Written by: Narration: Graphic Design: Mike Weimerskirch Mike Weimerskirch Mike Weimerskirch
Parabola - Equation Properties Examples Parabola Formula - Cuemath
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Equations of Parabolas - Kuta Software
= ?2y2 = 2y2 21) Vertex: 22) Vertex: = y2 + 10 = 3(x ? 4)2 + 2 Use the information provided to write the intercept form equation of each parabola 23) 24) = ?(x + 7)(x ? 4) = y2 + 20y + 103 Create your own worksheets like this one with Infinite Algebra 2 Free trial available at KutaSoftware com
QUADRATIC FUNCTIONS PARABOLAS AND PROBLEM SOLVING
2 5 Quadratic Functions Parabolas and Problem Solving 99 Graphs of quadratic functions For the quadratic functionf~x! 5 ax2 1 bx 1 c: The graph is a parabola with axis of symmetry x 5 2b 2a The parabola opensupward if a 0 downward if a 0 To ?nd the coordinates of the vertexset x 5 2b 2a Thenthey-coordinate is given by y 5 fS 2b 2a D
Searches related to parabole math equation PDF
The standard form of the equation of a parabolawith vertex at is as follows Ve rtical axis directrix: Horizontal axis directrix: The focus lies on the axis units (directed distance) from the vertex If the vertex is at the origin the equation takes one of the following forms Ve rtical axis Horizontal axis
What is the equation for a parabola?
Parabola is an important curve of the conic sections of the coordinate geometry. The general equation of a parabola is: y = a (x-h) 2 + k or x = a (y-k) 2 +h, where (h,k) denotes the vertex. The standard equation of a regular parabola is y 2 = 4ax. Some of the important terms below are helpful to understand the features and parts of a parabola.
What is the eccentricity of a parabola?
The eccentricity of a parabola is equal to 1. There are four standard equations of a parabola. The four standard forms are based on the axis and the orientation of the parabola. The transverse axis and the conjugate axis of each of these parabolas are different. The below image presents the four standard equations and forms of the parabola.
What is a parabola in a quadratic function?
Parabolas In Section 2.1, you learned that the graph of the quadratic function is a parabola that opens upward or downward. The following definition of a parabola is more general in the sense that it is independent of the orientation of the parabola.
Is a parabola symmetric with respect to its axis?
Note in Figure 10.10 that a parabola is symmetric with respect to its axis. Using the definition of a parabola, you can derive the following standard formof the equation of a parabola whose directrix is parallel to the -axis or to the -axis.
Introduction to conic sections
Author:
EduardOrtega
1 Introduction
A conic is a two-dimensional gure created by the intersection of a plane and a right circular cone. All conics can be written in terms of the following equation: Ax2+Bxy+Cy2+Dx+Ey+F= 0:
The four conics we'll explore in this text are parabolas, ellipses, circles, and hyperbolas. The equations for each of these conics can be written in a standard form, from which a lot about the given conic can be told without having to graph it. We'll study the standard forms and graphs of these four conics,1.1 General denition
A conic is the intersection of a plane and a right circular cone. The four basic types of conics are parabolas, ellipses, circles, and hyperbolas. Study the gures below to seehow a conic is geometrically dened.In anon-degenerate conicthe plane does not pass through the vertex of the cone.
When the plane does intersect the vertex of the cone, the resulting conic is called a degenerate conic. Degenerate conics include a point, a line, and two intersecting lines. The equation of every conic can be written in the following form: Ax2+Bxy+Cy2+Dx+Ey+F= 0:
1 This is the algebraic denition of a conic. Conics can be classied according to the coecients of this equation. Thediscriminantof the equation isB24AC. Assuming a conic is not degenerate, the following conditions hold true:1. IfB24AC <0 , the conic is acircle(ifB= 0 andA=B), or anellipse.
2. IfB24AC= 0, the conic is a parabola.
3. IfB24AC >0, the conic is ahyperbola.
Although there are many equations that describe a conic section, the following tablegives thestandard formequations for non-degenerate conics sections.Standard equation for non-degenerate conic section
circlex2+y2=a2ellipsex
2a 2+y2b2= 1parabolay
24ax= 0hyperbolax
2a 2y2b2= 11.2 problems
1. Is the following conic a parabola, an ellipse, a circle, or a hyperbola:3x2+y+2 =
0 ? It is a parabola.
2. Is the following conic a parabola, an ellipse, a circle, or a hyperbola: 2x2+ 3xy
4y2+ 2x3y+ 1 = 0 ? It is a hyperbola.
3. Is the following conic a parabola, an ellipse, a circle, or a hyperbola: 2x23y2= 0
? It is a hyperbola.4. Is the following conic a parabola, an ellipse, a circle, or a hyperbola:3x2+xy
2y2+ 4 = 0 ? It is an ellipse.
5. Is the following conic a parabola, an ellipse, a circle, or a hyperbola:x2= 0 ? It
is a degenerate conic.x= 0 is a line.6. Is the following conic a parabola, an ellipse, a circle, or a hyperbola:x2y2= 0 ?
It is a degenerate conic.x2y2= (xy)(x+y) = 0 are two lines that intersects.7. Is the following conic a parabola, an ellipse, a circle, or a hyperbola:x2+y2= 0
? It is a degenerate conic. The only point that satises the equationsx2+y2= 0 is (0;0). 21.3 Geometric denition
Let"be a positive number,eccentricity,`a line,directiceand a pointB,focus. The triple (";`;B) denes a conic section in the following way: A pointPis in the conic section dened by (";`;B) if jPBj=" jP`j jPBjstands for the distance from the pointPto the pointBandjP`jfor the minimal distance of the pointPto the line`. If the focusBdoes not belong to the directrice line`, the following conditions hold true:1. If 0< " <1 then conic is an ellipse.
2. If"= 1 then conic is an parabola.
3. If" >1 then conic is an hyperbola,
If the focusBdoes belong to the directrice line`, the following conditions hold true:1. If 0< " <1 then conic is a point.
2. If"= 1 then conic is a line.
3. If" >1 then conic are two lines that cross.
2 Non-degenerate conic sections
Given an eccentricity", a directrice line`and a focus pointBnot contain in`, we can dene a non-degenerate conic section. For simplicity we will assume that`is of the form x=LandB= (B;0), withL < B. We will see later that through translations and rotations we always can reduce to this situation. 3In this case given a pointP= (x;y) we have that
jPBj=p(xB)2+y2andjP`j=p(xL)2: Then the relationjPBj=" jP`jcan we written in the following way: p(xB)2+y2="p(xL)2:Then we have
(p(xB)2+y2)2= ("p(xL)2)2that is equivalent to(xB)2+y2="2(xL)2:So this is thegeneral equation of a conic section. Now we will study which type of conic
section is depending of the possible values of the eccentricity".2.1 Ellipse
We suppose that 0< " <1. First we compute the intersection of the conic section with thex-axis. To do that we have to replacey= 0 in the general equation of the conic section, so it follows the equation (xB)2="2(xL)2:This is equivalent to the equation
p(xB)2=p"2(xL)2;
4 so we have (xB) ="(xL); Here we encounter to possibilities: First suppose the equation (xB) ="(xL); that is equivalent to (1 +")x=B+"L; so the rst point that intersects thex-axis isx1=x=B+"L1+":Finally suppose the equation
(xB) = +"(xL); that is equivalent to (1")x=B"L; so the second point that intersects thex-axis isx2=x=B"L1":A simple calculation yields thatx1< x2.denitions
centerx=x1+x22 major axisa=x2x12 minor axisb=ap1"2With this denitions on hand we can rewrite the general equation in the following way(xx)2a 2+y2b2= 1;that we callthe standard equation of the ellipse.
Conversely,(xx)2a
2+y2b2= 1eccentricity"=q1b2a
2directriceL= xa"
focusB= x"a5 From the standard equation of the ellipse one can observe that the ellipse is symmetric with respect to the vertical linex= x. Therefore if we dene B2= x+a"
and L2= x +a" we have that the triple given by eccentricity", focus pointB2= (B2;0) and the directrice line`2given byx=L2, determines the same conic section as (";B;`). Thus,B1=BandB2are called the two focus points of the ellipse.Now given the two focus of the ellipseB1andB2we can give an alternative geometric
description, in the following way: An ellipse is the set of points such that the sum of the distances from any point on the ellipse toB1andB2is constant and equal to 2a, that is jPB1j+jPB2j= 2a6
2.1.1 Examples
1. Find the equation of the ellipse with eccentricity"= 1=3, directrice linex=1
and focusB= (1;0). Then according the formulas we have x1=1 + 1=3(1)1 + 1=3=2=34=3= 1=2x2=11=3(1)11=3=4=32=3= 2
and hence the center of the ellipse is x=1=2 + 22 = 5=4; and a=21=22 = 3=4b= 3=4p1(1=3)2= 3=4p8=9 =p2 2Therefore the equation is
(x5=4)2(3=4)2+y2( p2 2 )2= 1; so16(x5=4)29 + 2y2= 12. LetB1= (1;0) andB2= (3;0) be two points in the plane. We want to give the equation of the ellipse such pointsPsatisfy jPB1j+jPB2j= 6:7
First observe that the formulas say that 2a= 6, and hencea= 3. The center of the ellipse is the mid-point betweenB1andB2that is x= 2. We need now to calculate the eccentricity, that from the above formulas comes from the relation jB1B2j= 2a";
so we have that 4 = 23", and follows that"= 2=3. Finally, we have that b=ap1"2, sob= 3p5=9 =p5. Therefore the equation of the ellipse is (x2)29 +y25 = 12.2 Parabel We suppose that"= 1, that translates as the condition jPBj=" jP`j; that are the pointsPin the plane that are at the same distance from the focusBas from the directrice`. Then the general equation of the conic section reduces to (xB)2+y2= (xL)2; and we can write it as y2= (xL)2(xB)2=x22xL+L2x2+ 2xBB2= 2(BL)x+ (L2B2);
that isy2= 2(BL)x+ (L2B2)If we want to nd the intersection of the conic section with thex-axis, we have to
replacey= 0 in the above equation. So we have0 = 2(BL)x+ (L2B2)
that is2(LB)x= (L2B2) = (LB)(L+B)
so after cancel out some the (LB) term we have thatx1=x=L+B2
8 that we can thevertexof the parabola.2.3 Hyperbola We suppose that" >1. First we compute the intersection of the conic section with the x-axis. To do that we have to replacey= 0 in the general equation of the conic section, so it follows the equation (xB)2="2(xL)2:This is equivalent to the equation
p(xB)2=p"2(xL)2;
so we have (xB) ="(xL); Here we encounter to possibilities: First suppose the equation (xB) ="(xL); that is equivalent to (1 +")x=B+"L; so the rst point that intersects thex-axis isx1=x=B+"L1+":Finally suppose the equation
(xB) = +"(xL); 9 that is equivalent to (1")x=B"L; so the second point that intersects thex-axis isx2=x=B"L1":A simple calculation yields thatx1> x2. Observe that this is the opposite that
happens in the ellipse situation.denitions centerx=x1+x22 major axisa=x1x22 minor axisb=ap"21With this denitions on hand we can rewrite the general equation in the following
way(xx)2a 2y2b2= 1;that we callthe standard equation of the hyperbola.Conversely,
(xx)2a 2y2b2= 1eccentricity"=q1 +
b2a2directriceL= x+a"
focusB= x+"a10 From the standard equation of the hyperbola one can observe that the hyperbola is symmetric with respect to the vertical linex= x. Therefore if we dene B2= x"aand L2= xa"
we have that the triple given by eccentricity", focus pointB2= (B2;0) and the directrice line`2given byx=L2, determines the same conic section as (";B;`). Thus,B1=B andB2are called the two focus points of the hyperbola. Now given the two focus of the hyperbolaB1andB2we can give an alternative geometric description, in the following way: An hyperbola is the set of points such that the dierence of the distances from any point on the ellipse toB1andB2is constant and equal to 2a, that is jPB1j jPB2j= 2aorjPB2j jPB1j= 2a:2.3.1 Examples
1. Find the equation of the hyperbola with eccentricity"= 2, directrice linex=1
and focusB= (1;0). Then according the formulas we have x1=1 + 2(1)1 + 2
=13 x2=12(1)12=31=3 and hence the center of the hyperbola is x=1=332 =5=3; and a=1=3(3)2 = 4=3b= 4=3p221 = 4=3p3 =
4p3 11Therefore the equation is
(x+ 5=3)2(4=3)2y2( 4p3 )2= 1; so9(x+5=3)216 3y216 = 12. LetB1= (1;0) andB2= (3;0) be two points in the plane. We want to give the equation of the hyperbola such pointsPsatisfy jPB1j jPB2j= 6 orjPB2j jPB1j= 6:First observe that the formulas say that 2a= 6, and hencea= 3. The center of
the ellipse is the mid-point betweenB1andB2that is x= 2. We need now to calculate the eccentricity, that from the above formulas comes from the relation jB1B2j=2a"
so we have that 4 = 23", and follows that"= 3=2. Finally, we have thatb= ap"
21, sob= 3p5=4 =3p5
2 . Therefore the equation of the hyperbola is(x2)29 4y245 = 1123 Change of coordinates
In the above section we have supposed that the directrice line is parallel to they-axis, i.e.,x=Land the focus is over thexaxis, i.e.B= (B;0), but what happens with the general situation where we have any given line and point? Change of coordinates!3.1 translation Atranslation to a point(a;b) is a change of coordinates (x;y) to a new coordinates (x;y) in such a way x=xaand y = yb: Roughly speaking, a translation moves the origin to the point (a;b).13 We can reverse the change of coordinates from the new coordinates (x;y) to the old ones: x= x+aand y = y + b:3.1.1 Examples
1. We consider the translation to the point (1;2). Then:(x;y)-coordinates(x;y)-coordinates(0;0)(1;2)(1;2)(0;0)y=xy+ 2 = x+ 1y= x1x
2+y2= 1(x+ 1)2+ (y+ 2)2= 114
2. We want to nd the equation of the ellipse that has eccentricity"= 1=2, directrice
linex= 1 and focus (3;2).Observe that if we make a translation to the point (2;2) we have the following(x;y)-coordinates(x;y)-coordinates(3;2)(1;0)x= 1x+ 2 = 1x=1So now we can construct the ellipse with eccentricity"= 1=2, directrice line
x=1 =Land focus (1;0), soB= 1. According the formulas we have that x1=1 + 1=2(1)1 + 1=2=1=23=2= 1=3
and x2=11=2(1)11=2=3=21=2= 3:
thus x=1=3 + 32 =10=32 = 5=3 a=31=32 =8=32 = 4=3 and b= 4=3p1(1=2)2= 4=3p3=4 =2p3 Therefore the equation of the ellipse in the (x;y) coordinates is (x5=3)2(4=3)2+y2( 2p3 )2= 1 15 that we can rewrite as (x5=3)216=9+y24=3= 1:Finally we return to the old coordinates (x;y), using that x=x2 and y = y2:So replacing this to the equation we have
((x2)5=3)216=9+(y2)24=3= 1: that is(x11=3)216=9+(y2)24=3= 1:163.2 rotation
Arotation with angleis a change of coordinates (x;y) to a new coordinates (x;y) in such a wayx=xcos+ysinand y =xsin+ ycos:We can reverse the change of coordinates from the new coordinates (x;y) to the old
ones: x= xcosysinand y = xsin+ ycos:3.2.1 Examples
1. We consider the a rotation of 45
. Then:(x;y)-coordinates(x;y)-coordinates(0;0)(0;0)(1;1)( p2;0)y=x1( p2 2 x+p2 2 y) =(p2 2 xp2 2 y)1x=1p2 172. We want to nd the equation of the parabola with directrice liney=x1 and
focus (1;1).18Observe that if we make a translation of 45
in the new coordinates (x;y) the directrice line has equation x=1p2 and the focus (p2;0). Then we can write the equation of the parabola y2= 2(p2(1p2 ))x+ ((1p2 )2)(p2) 2=3p2 x32 Finally, making the change of coordinate to the old coordinates (x;y), we have that (p2 2 x+p2 2 y)2=3p2 (p2 2 x+p2 2 y)32 so it follows that12 x2+12 y2xy=32 x+32 y32 and hence the nal equation isx2+y22xy3x3y+ 3 = 019
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