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Eccentricity =1 in parabola

Learning objectives:

This article explains one of the conic sections The PARABOLA  in detail.

A brief  introduction on the  conics sections and types of conic sections, axis of symmetry, eccentricity can be reviewed  in the below  article Conic section introduction

Following are the concepts covered and features of the current article

  • The Focus of a parabola & Directrix .
  • Latus rectum of a parabola .
  • The relation between the focus, directrix, and the vertex of a parabola.
  • Different forms of parabola equations.
  • Finding the equation of a parabola with focus and directrix.

 

The Focus of a parabola & Directrix:

Parabola Focus is a point that lies on the axis of symmetry of the parabola. Parabola encloses the Focus point.

Directrix is a line perpendicular to the axis of symmetry and at a focal distance “a” from the vertex. The vertex of the parabola lies exactly in the middle of the focus point and the directrix line.

Focus is a fixed point, wherefrom the length of a line drawn starting from it, to any point on the parabola, and the perpendicular line drawn from the point on the parabola to the directrix will be the same.

Let us consider a parabola with a point “P” on it. Let the Parabola focus be located at a point “F”.A perpendicular line is drawn from the point  P to the directrix line, and the intersection point is D.

Moreover,the interesting fact is that, the FP will be equal to PD as shown in the picture below.

 

 

Eccentricity =1 in parabola
FP=PD in a parabola

The distance between the Parabola focus and any point on parabola FP, will be the same as the perpendicular distance between the point on the parabola and the directrix PD. FP=PD.

The interesting fact is that any ray incident inside the parabola inside surface converges at the focus F.This is the principle used in the dish antennas, where the receptor is located at the focus of the dish antenna.

The Similar way, any signal source emerging from the focus of the parabola will incident on the parabola and they form parallel lines moving away from the parabola. This principle is generally used for the broadcasting of the signals.

 

Latus rectum of a parabola:

The Latus rectum of a parabola is the perpendicular line passing through focus and intersects the parabola at the two ends say L L’.

Different forms of parabolic equations:

Parabola Equations with vertex at (0,0).

1)y^{2}=\pm 4ax [ -4ax is left  opening and 4ax is right  opening parabola]

2) x^{2}=\pm 4ay [4ay is up opening  and -4ay is down opening parabola]

Memorizing the opening of the parabola will be quite confusing at times. The easy way to remember is to observe the non-squared variable.

There is an easy way to figure out the opening of a  parabola, just by looking at the non- squared variable.

Let us say we have the parabola equation y^{2}=4ax . The squared variable is Y and the non-squared variable term is the  “x”.So the graph will be either \subset or  \supset shaped. This looks more or less like the alphabet “x”.

Therefore y^{2}=4ax graph looks like a \subset or  \supset .

Similar way, if we have x^2=4ay, we observe the non- squared variable term “y” So the graph will be either a “U” shaped or an inverted “U” shaped.The  alphabet “y”  has a “u” shape in it .

 

Graph of the parabola equation y^{2}=4ax opening right

 

y^2 = 4ax

 

i) The axis of symmetry is Y=0 ( the x-axis)

ii)Vertex of parabola is the  origin (0,0)

iii)Coordinates of  focus F is (a,0)

iv) Length of latus rectum is 4a.

v)Equation of  latus rectum is  x =a.

vi) Equation of  directrix is x =-a.

vii)End coordinates of latus rectum are (a,2a) and (a,-2a)

viii)Tangent at the  vertex (0,0) is the  X=0 ( the  y- axis).

ix) The latus rectum length is “4a”that is 4 times the focal length “a”.

 

Graph of the parabola equation y^{2}=-4ax

Left opening parabola
Left Opening Parabola

i) The axis of symmetry is y=0 ( the x-axis)

ii)Vertex of parabola is the  origin (0,0)

iii)Coordinates of  focus F is (-a,0)

iv) Length of latus rectum is LL’=4a.

v)Equation of  latus rectum is  x =-a.

vi) Equation of  directrix is x =a.

vii)End coordinates id latus rectum are (-a,2a) and (-a,-2a)

viii)Tangent at the  vertex (0,0) is the  X=0 ( the  y- axis).

ix) The latus rectum length is “4a” that is  4 times thefocal length “a”.

 

 

 

Graph of the parabola equation x^{2}=4ay opening Up.

 

i)The axis of symmetry is x=0 ( the y-axis) up-facing Parabola

ii)Vertex of parabola is the  origin (0,0)

iii)Coordinates of  focus (0,a)

iv) Length of latus rectum LL’= 4a.

v)Equation of  latus rectum is  y =a.

vi) Equation of  directrix is y=-a.

vii)End coordinates of latus rectum are (2a,a) and (-2a,a)

viii)Tangent at the  vertex (0,0) is the  y=0 ( the  x- axis).

ix) The latus rectum length is “4a” that is  4 times the focal length “a”.

 

 

 

Graph of the parabola equation x^{2}=-4ay opening Down.

Down parabola
X^2 =-4ay down facing parabola

 

i)The axis of symmetry is x=0 ( the y-axis)

ii)Vertex of parabola is the  origin (0,0)

iii)Coordinates of  focus (0,-a)

iv) Length of latus rectum LL’= 4a.

v)Equation of  latus rectum is  y =-a.

vi) Equation of  directrix is y=a.

vii)End coordinates of latus rectum are (-2a,-a) and (2a,-a)

viii)Tangent at the  vertex (0,0) is the  y=0 ( the  x- axis).

ix) The latus rectum length is “4a” that is  4 times the focal length “a”.

 

 

Parabola Equations with vertex at (h,k) other than the origin (0,0).

Parabola equation form \left( y-k\right) ^{2}=4a\left( x-h\right) also called the  vertex form of  parabola, had the  center  at  (h,k).

As a result,There will be a minor change in the coordinates of focus, Latus rectum, and the directrix.

The focus and Directrix co-ordinates are shifted by the (h,k). All we need is to add the  (h,k) to the parent parabolic equations discussed above.

Below is the summary table of parabolic equations with different orientations and vertex at the origin and at  (h,k).

Parabola GraphParabola EquationVertex point
Axis Of SymmetryFocus pointDirectrixLatus rectum
summary table y^2 =4axy^2 = 4ax(0,0) y=0
(X-Axis)
( a,0)x=-a(a,2a)
(a,-2a)
( y-k)^2 = 4a (x-h)(y-k)^2=
4a(x-h)
(h,k)y=k(a+h, k)x=-a+h(a+h,2a+k)
( a+h,-2a+h)
y^ =-4axy^2 = -4ax(0,0)y=0 (X-Axis)( -a,0)x=a(-a,2a)
(-a,-2a)
 ( y-k)^2 =-4a (x-h)(y-k)^2=
-4a(x-h)
(h,k)y=k(-a+h,k)x= a+h(-a+h,2a+k)
(-a+h,-2a+k)

Finding the equation of a parabola with focus and directrix:

The key step in making the parabola equation is observing the orientation of the focus coordinate, whether the focus is on right /left/ above or below the directrix line.

Let’s say we are given the focus coordinates as (P, Q) and a directrix line equation. The following are a series of steps to find the parabola Equation.

i)Firstly, Identify the orientation of the focus point , with respect to the directrix line. Check the directrix line whether it is a  horizontal or a vertical line.

ii)Find the vertex (h,k) by finding the mid-point of the focus and the directrix line.

iii)Get the focal length “a” from the vertex calculated to the focal point.

iv)If the directrix of the form Y= Constant number,( A horizontal line) then the parabola equation would be

1)(X-h)^2 = 4a(Y-k) (If the  focus point is above the directrix  line)

2)(X-h)^2=-4a(Y-k)(If the  focus point is below the directrix  line )

v): If the directrix is of the form  X= constant number,( A vertical line) then the parabola equation would be

1)(Y-k)^2 = 4a(X-h) (If the  focus point is to the right of  the directrix  line)

2)(Y-k)^2=-4a(X-h) (If the  focus point is to the left of  the directrix  line)

 

Examples:

Given the focus of the parabola is ( 2, 1)  and the directrix line is  X= -2. Find the equation of the parabola.

Solution:

 

Step1: “Firstly, Identify the orientation of the focus point , with respect to the directrix line. Check the directrix line whether it is a  horizontal or a vertical line.

The focus (  2,1)  is located on the right side of the directrix line x =-2. The line x=-2 is a vertical line.

 

Step2: “Find the vertex (h,k) by finding the mid-point of the focus and the directrix line.

The “Y” coordinate remains the same y =1 and the “x” coordinate is the average of the -2 and 2 which is 0. Hence the Vertex is  (0,1)

 

Step3: “Get the focal length “a” from the  vertex and the  focal point.”

Focus is (2,1) and the  vertex is (0,1).Hence the focal length is 2.

 

Step4: If the directrix is of the form  X= constant number, then the parabola equation would be (Y-k)^2 = 4a(X-h) (If the  focus point is to the right of  the directrix  line)

Here the focus point is to the right side of  the directrix line. Therefore ,the parabola equation with vertex (0,1) and a=2 would be.

(Y-1)^2 =4*2(X-0)

= (Y-1)^2 =8(X).

 

(y-1)^2 = 8x example parabola

 

More Examples:

Find the parabola equation given the graph {Directrix y=5 and focus(4,-7)}