Introduction Parabola Ellipse

Conic Sections

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54–Elementary Analysis 2

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Introduction Parabola Ellipse

Introduction

A conic section is a plane figure formed when a right-circular cone is cut by a plane.Below are samples of conic sections.

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Introduction Parabola Ellipse

Parabola

The set of all points in the plane whose distance from a fixed point (focus) is equalto its distance from a fixed line (directrix).

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Introduction Parabola Ellipse

Equation of a Parabola

Consider the directix and the focus situated this way:

A point P = (x,y) is on the parabola ifdistance of P to F = distance of P to D√(x−p)2 + (y−0)2 = ∣∣x− (−p)

∣∣√(x−p)2 +y2 = ∣∣x+p

∣∣(x−p)2 +y2 = (x+p)2

x2 −2px+p2 +y2 = x2 +2px+p2

y2 = 4px

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Introduction Parabola Ellipse

Equation of a Parabola

y2 = 4px

focus: (p,0)

directrix: x =−p

The line through the focus that is perpendicular to the directrix is called axisof symmetry or simply, axis.

The intersection of the axis and the parabola is the vertex Its distance fromthe focus is p, which is its distance from the directrix.

The line segment parallel to the directrix that contains the focus is calledlatus rectum.

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Introduction Parabola Ellipse

Equation of a Parabola

In summary, we have

Equation Focus Directrix +4p −4p

y2 =±4px (±p,0) x =∓p

x2 =±4py (±p,0) y =∓p

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Introduction Parabola Ellipse

Equation of a Parabola

Example.

Sketch the graph of y = 12x2.

Solution:

The equation can be written as y = 4(3)x.

Here, p = 3, and the parabola opens up.

The vertex is at the origin.

The focus is at (0,3).

The endpoints of the latus rectum are at(−6,3) and (6,3).

Hence, the graph is shown on the right.

Additionally, the directrix is the liney =−3.

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Introduction Parabola Ellipse

Equation of a Parabola

Example.

Sketch the graph of 4x+3y2 = 0.

Solution:

The equation can be written as

4x =−3y2, or 4(− 1

3

)x = y2.

Here, p = 13 , and the parabola opens to

the left.

The vertex is at the origin.

The focus is at(− 1

3 ,0).

The endpoints of the latus rectum are at(− 1

3 , 23

)and

(− 1

3 ,− 23

).

Hence, the graph is shown on the right.

Additionally, the directrix is the linex = 1

3 .

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Introduction Parabola Ellipse

Equation of a Parabola

In general, if the vertex of a parabola is at the point (h,k), then it has either of theform

(y−k)2 =±4p(x−h) (x−h)2 =±4p(y−k)

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Introduction Parabola Ellipse

Equation of a Parabola

Parabola opening to the right

(y−k)2 = 4p(x−h) (p > 0)

vertex : (h,k)

focus : (h+p,k)

latus rectum :(h+p,k±2p)

directrix : x = h−p

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Introduction Parabola Ellipse

Equation of a Parabola

Parabola opening to the left

(y−k)2 =−4p(x−h) (p > 0)

vertex : (h,k)

focus : (h−p,k)

latus rectum :(h−p,k±2p)

directrix : x = h+p

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Introduction Parabola Ellipse

Equation of a Parabola

Parabola opening upward

(x−h)2 = 4p(y−k) (p > 0)

vertex : (h,k)

focus : (h,k+p)

latus rectum :(h±2p,k+p)

directrix : y = k−p

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Introduction Parabola Ellipse

Equation of a Parabola

Parabola opening downward

(x−h)2 =−4p(y−k) (p > 0)

vertex : (h,k)

focus : (h,k−p)

latus rectum :(h±2p,k−p)

directrix : y = k+p

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Introduction Parabola Ellipse

Equation of a Parabola

Example.

Sketch the graph of y2 +4x+8y = 0. Also, identify the focus, endpoints of latusrectum, and the directrix.

Solution:

The equation can be written asy2 +8y+16 =−4x+16, or(y− (−4)

)2 =−4(1)(x−4).

Here, p = 1, and the parabola opens tothe left.

The vertex is at (4,−4).

The focus is at (3,−4).

The endpoints of the latus rectum are at(3,−2) and (3,−6).

Hence, the graph is shown on the right.

Also, the directrix is the line x = 5.

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Introduction Parabola Ellipse

Equation of a Parabola

Example

Find the equation of the parabola given that the focus is at (3,2) and the directrix isthe line x = 1.

Solution:

We plot the focus and draw the directrix.

Thus, the vertex is at (3,3), and theparabola opens downwards.

Hence, p = 1 =⇒ 4p = 4.

Thus, the equation is

(x−3)2 =−4(y−3).

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Ellipse

An ellipse is a set of points in the plane whose distances from two fixed points(focuses/foci) sum up to a constant.

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Equation of an Ellipse

Foci : F1 = (c,0) and F2 = (−c,0)

Vertices : V1 = (a,0) and V2 = (−a,0)

Endpoints of minor axis: (0,b) and (0,−b)

Point P = (x,y) is on the ellipse if

distance of P to F1 + distance of P to F2 = constant2a√(x+ c)2 +y2 +

√(x− c)2 +y2 = 2a

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Introduction Parabola Ellipse

Equation of an Ellipse

√(x+ c)2 +y2 = 2a−

√(x− c)2 +y2

x2 +2cx+ c2 +y2 = 4a2 −4a√

(x− c)2 +y2 +x2 −2cx+ c2 +y2

4a√

(x− c)2 +y2 = 4a2 −4cx

a2(x2 −2cx+ c2 +y2

)= a4 −2a2cx+ c2x2

a2x2 −2a2cx+a2c2 +a2y2 = a4 −2a2cx+ c2x2(a2 − c2

)x2 +a2y2 = a2

(a2 − c2

)x2

a2+ y2

a2 − c2= 1

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Introduction Parabola Ellipse

Equation of an Ellipse

x2

a2+ y2

a2 − c2= 1

Notice that b2 = a2 − c2.

Thus, we have

x2

a2+ y2

b2= 1.

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Equation of an Ellipse

Ellipse with horizontal major axis

An ellipse with horizontal major axis and center at (0,0) has the form

x2

a2+ y2

b2= 1

where (a > b)

vertices : (±a,0)

length of major axis = 2a

minor axis: (0,±b)

length of minor axis = 2b

foci : (±c,0)

where a2 −b2 = c2

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Introduction Parabola Ellipse

Equation of an Ellipse

Remark. An ellipse with vertical major axis

An ellipse with vertical major axis and center at (0,0) has the form

x2

a2+ y2

b2= 1

where (b > a)

vertices : (0,±b)

length of major axis = 2b

minor axis: (±a,0)

length of minor axis = 2a

foci : (0,±c)

where b2 −a2 = c2

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Introduction Parabola Ellipse

Ellipse

Example.

Sketch the graph ofx2

25+ y2

9= 1.

Solution. The equation can be written asx2

52+ y2

32= 1.

Hence, a = 5, b = 3, and major axis is horizontal, with center at (0,0).

vertices : V1(5,0), V2(−5,0)

minor axis : B1(0,3), B2(0,−3)

c2 = a2 −b2= 52 −32= 16= 42

foci : F1(4,0), F2(−4,0)

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Introduction Parabola Ellipse

Ellipse

Example.

Sketch the graph of 9x2 +4y2 = 36.

Solution. The equation can be written asx2

22+ y2

32= 1.

Hence, a = 2, b = 3, and major axis is vertical, with center at (0,0).

vertices : V1(0,3), V2(0,−3)

minor axis : B1(2,0), B2(−2,0)

c2 = b2 −a2= 32 −22= 5=(p

5)2

foci : F1(0,p

5), F2(0,−p5)

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Introduction Parabola Ellipse

Equation of an Ellipse

In general, an ellipse centered at (h,k) has the equation given by

(x−h)2

a2+ (y−k)2

b2= 1

Major Axis Vertices Endpoints of the Minor Axis

(a > b) horizontal (h±a,k) (h,k±b)

(b > a) vertical (h,k±b) (h±a,k)

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Introduction Parabola Ellipse

Ellipse

Example.

Sketch the graph of 25x2 +y2 −4y−21 = 0.

Solution. The equation can be written as

25x2 +y2 −4y+4 = 21+4 =⇒ 25x2 + (y−2)2 = 25 =⇒ (x−0)2

12+ (y−2)2

52= 1

Hence, a = 5, b = 1, and major axis is vertical, with center at (0,2).

vertices : V1(0,7), V2(0,−3)

minor axis : B1(1,2), B2(−1,2)

c2 = a2 −b2= 52 −12= 24=p24

2

foci : F1(0,2+p24),

F2(0,2−p24)

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Introduction Parabola Ellipse

Equation of an Ellipse

Example

Find the equation of the ellipse with one vertex at (3,2) and its minor axis has oneendpoint at (−1,−1)

Solution:

Center : (−1,2)

a = 4

b = 3

The major axis ishorizontal.

Thus, the equation is

(x+1)2

16+ (y−2)2

9= 1.

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