math 54 lecture 6 - conic sections (parabola and ellipse)
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Math 54 Lecture 6 - Conic Sections (Parabola and Ellipse)TRANSCRIPT
Introduction Parabola Ellipse
Conic Sections
Institute of Mathematics, University of the Philippines Diliman
Mathematics 54–Elementary Analysis 2
Conic Sections 1/ 26
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.
Conic Sections 2/ 26
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).
Conic Sections 3/ 26
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
Conic Sections 4/ 26
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.
Conic Sections 5/ 26
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
Conic Sections 6/ 26
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.
Conic Sections 7/ 26
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 .
Conic Sections 8/ 26
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)
Conic Sections 9/ 26
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
Conic Sections 10/ 26
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
Conic Sections 11/ 26
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
Conic Sections 12/ 26
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
Conic Sections 13/ 26
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.
Conic Sections 14/ 26
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).
Conic Sections 15/ 26
Introduction Parabola Ellipse
Ellipse
An ellipse is a set of points in the plane whose distances from two fixed points(focuses/foci) sum up to a constant.
Conic Sections 16/ 26
Introduction Parabola Ellipse
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
Conic Sections 17/ 26
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
Conic Sections 18/ 26
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.
Conic Sections 19/ 26
Introduction Parabola Ellipse
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
Conic Sections 20/ 26
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
Conic Sections 21/ 26
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)
Conic Sections 22/ 26
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)
Conic Sections 23/ 26
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)
Conic Sections 24/ 26
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)
Conic Sections 25/ 26
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.
Conic Sections 26/ 26