AN 06 00 |
- Outlines
AN 06 01 |
- Defintion of Locus of parabola
AN 06 02 |
- Equations of parabola
AN 06 03 |
- Equations in polar form
AN 06 04 |
- Equations in parametric form
AN 06 05 |
- Example : Find equation for locus when PF = F to directrix
AN 06 06 |
- Example : If y = x^2 + 2*x + 1, find coordinates of focus
AN 06 07 |
- Example : Sketch parabola using ruler
AN 06 08 |
- Example : Prove that R = csc(A/2)^2 is parabola
AN 06 09 |
- Example : Prove that R = sec(A/2)^2 is parabola
AN 06 10 |
- Example : If y = x^2, draw tangent to curve when x=3
AN 06 11 |
- Find coordinate of focus for y = x^2 - 6*x + 8
AN 06 12 |
- Find equation of directrix of y = x^2
AN 06 13 |
- Prove the locus of parabola y-k = a*(x-h)^2
AN 06 14 |
- Convert y = a*x^2 + b*x + c to polar form
AN 06 15 |
- Draw tangent to parabola by law of reflextion
AN 06 16 |
- Convert R = D/(1-sin(A)) to y = F(x)
AN 06 17 |
- Elliminate x*y term in F(x,y)
AN 06 18 |
- Compare y+D/2 = x^2/(2*D) with R = D/(1-sin(A))
AN 06 19 |
- Example : Study parabola of y = x^2 - 6*x + 8
AN 06 20 |
- Example : Applications of parabola
AN 06 21 |
- Example : Three points define a parabola
AN 06 22 |
- Subjects of parabola on internet
AN 06 23 |
- References
AN 06 00 |
- Outlines
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Answers
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AN 06 01. Defintion of locus of a parabola
Defintion
Figure 1
- One fixed point F and and a fixed line. A moving point P(x,y).
- When P moves so that PF = point P to fixed line, what is the locus ?
- Answer : The locus of point P is a parabola.
- Defintion
- Where F is the focus. The fixed line is the directrix.
- The focus F is on the major axis which is perpendicular to direcrix.
- The equation is (y+D/2) = x^2/(2*D).
- The origin is at F and (0,-D/2) is the vertex.
- Focus F to directrix is D.
- Focus to vertex V is D/2.
Go to Begin
AN 06 02. Equations of Parabola
[Standard Form]
- Equation in vertex form : y-k = a*(x-h)^2.
- Where V(h,k) is the vertex. The values a is 1/(2*D).
- Major axis and openning
- If a is positive, it opens upward.
- If a is negative, it opens downward.
- The vertex
- The point is minimum if a is positive.
- The point is maximum if a is negative.
[Implicit Form without x*y]
- Equation : F(x,y) = A*x^2 + C*y^2 + D*x + E*y + F = 0.
- Focus is on line parallel to x-axis or y-axis.
- A or C is zero.
- Change to standard form by using completing the square.
- Hence we can find the vertex (h,k).
- Hence we can find postion of F and equation of directrix.
[Implicit Form with term x*y]
- Equation : F(x,y) = A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0.
- B^2 - 4*A*C = 0 if it is a parabola.
- Focus is not on line parallel to x-axis or y-axis.
- How to find principal axis ?
- We must elliminate x*y by rotating an angle.
- The new coefficients A' C' D' E' F' can be obtained with B' = 0.
- Then we can find the principal axis.
Go to Begin
AN 06 03. Equations in Polar Form :
Figure 2
[Function 1]
- Function : R = D/(1-sin(A)).
- D is focus to directrix line.
- If F is origin, then equation of directrix is y = -D.
- A is angle making with x-axis.
- Focus are on y-axis and it opens upward.
[Function 2]
- Function : R = D/(1+sin(A)).
- D is focus to directrix line.
- If F is origin, then equation of directrix is y = D.
- A is angle making with x-axis.
- Focus is on y-axis and it opens downward.
[Function 3]
- Function : R = D/(1-cos(A)).
- D is focus to directrix line.
- If F is origin, then equation of directrix is x = -D.
- A is angle making with x-axis.
- Focus is on x-axis and it opens to the left.
[Function 4]
- Function : R = D/(1+cos(A)).
- D is focus to directrix line.
- If F is origin, then equation of directrix is x = D.
- A is angle making with x-axis.
- Focus is on x-axis and it opens to the right.
Go to Begin
AN 06 04. Equation in parametric form
It is not often useful as ellipse
- x = F(t).
- y = G(t).
- y-k = a*(x-h)^2
- (h,k) is the vertex.
Go to Begin
AN 06 05. Example : Fixed point F(0,2) and fixed line at y = -2. A moving point is P(x,y)
If PF = F to fixed line, find the equation of the motion.
[Method 1] By defintion of standard equation
- The motion is a parabola and focus is on the y-axis.
- Distance between F and directrix is D = 4.
- The vertex is between F and directrix. It is at (0,0).
- The equation is (y-0) = (x-0)^2/(2*D).
- The equation is y = x^2/8
[Method 2] By using distance formula
- Since PF = F to directrix and D = 4.
- Let origin be at vertex. Focus at (0,2).
- Distance between P and F is Sqr(x^2+(y-2)^2).
- Distance between P and directrix is y + D/2.
- Hence Sqr(x^2 + (y-2)^2) = y + D/2.
- Square both sides we have
- x^2 + y^2 - 4*y + 4 = (y + D/2)^2.
- x^2 + y^2 - 4*y + 4 = y^2 + 4*y + 4.
- x^2 = 8*y.
- Hence y = x^2/8.
Go to Begin
AN 06 06. Example : If y = x^2 + 2*x + 1, find coordinates of focus
- If y = a*x^2 + B*x + c, the vertex is at x = -b/(2*a).
- The focus to directrix is D = 1/(2*a).
- The focus is at x = xv and y = yv+D/2 where xv=-1 and yv=0.
- The vertex is at x = -1 and y = 0. Principal axis is x = -1.
- The focus is at F(-1,1/4).
Go to Begin
AN 06 07. Example : Sketch a parabola using ruler :
Diagram
[Method 1]
- Draw a point F as focus.
- Draw a horizontal line as directrix.
- Draw a line QR perpendicular to directrix.
- Q is a point on the directrix.
- Join Q and F. Bisect QF.
- Bisector meets line QR at P.
- P is a point on the locus because PF = PQ.
- Repeat above to get more point on locus.
[Method 2] Compute (x,y) using equation in rectangular form
[Method 3] Compute (R,A) using polar function
Go to Begin
AN 06 08. Example : Prove that R = csc(A/2)^2 is parabola
Proof
- Since csc(B/2) = 1/sin(B/2) and sin(B/2) = Sqr((1-cos(B))/2).
- Hence R = csc(A/2)^2 = 1/sin(A/2)^2.
- Hence R = 2/(1-cos(A)).
- It is parabola.
- D = 2.
- If F is origin, then equation of directrix is x = -2.
- If F is origin, then vertex is at (-1,0).
- It opens to left.
- Directrix is at left.
Go to Begin
Q9. Example : Prove that R = sec(A/2)^2 is parabola
Proof
- Since sec(B/2) = 1/cos(B/2) and cos(B/2) = Sqr((1+cos(B))/2).
- Hence R = sec(A/2)^2 = 1/cos(A/2)^2.
- Hence R = 2/(1+cos(A)).
- It is parabola.
- D = 2.
- If F is origin, then equation of directrix is x = 2.
- If F is origin, then vertex is at (1,0).
- It opens to right.
- Directrix is at right.
Go to Begin
AN 06 10. Example : If y = x^2 draw tangent to curve at x=3
Solution
- Slope of curve is y' = 2*x.
- Slope at x = 3 is m = 6.
- Equation of tangent is y = m*x + n.
- When x = 3 and y = 9.
- Hence 9 = 6*3 + n. and n = - 9
- Equation of tangent is y = 6*x - 9.
Go to Begin
Q11. Find coordinate of focus for y = x^2 - 6*x + 8
Defintion
- Focal point is on principal axis
- It has a distance D from the directrix
- It has a distance D/2 from the vertex
- D = 1/(2*a) if y = a*x^2 + b*x + c
Solution
- Distance from focus to directrix : D = 1/(2*a) = 1/2
- Vertex
- xv = -b/(2*a) = -(-6)/(2*1) = 3
- yv = (3)^2 - 6*3 + 8 = 9 - 18 + 8 = -1
- Focus
- xf = xv = 3
- yf = yv + D/2 = -1 + (1/2)/2 = -0.75
Go to Begin
AN 06 12. Find equation of directrix of y = x^2
Defintion
- Directrix is perpendicualr to principal axis
- It has a distance D from the focus
- It has a distance D/2 from the vertex
- D = 1/(2*a) if y = a*x^2 + b*x + c
Solution
- D = 1/2
- Vertex of parabola is at (0,0) or xv = 0 and yv = 0
- Focus is at xf = xv and yf = yv + D/2
- Equation of directrix is y = yv - D/2 = 0 - (1/2)/2 = 0.25
Go to Begin
AN 06 13. Example : Prove that parabol of ellipse is (y-k)=a*(x-h)^2
- Point P(x,y) to focal point F(0,0) is Sqr(x^2+y^2).
- Assume origin at F and D = focus to directrix.
- The vertex is at (0,-D/2).
- Distance of P to directix is y + D.
- By definition of locus of parabola : PF = P to directrix.
- Hence Sqr(x^2+y^2) = y+D.
- Squre both sides we have
- x^2 + y^2 = (y+D)^2.
- x^2 + y^2 = y^2 + 2*D*y + D^2.
- 2*D*y + D^2 = x^2.
- y + D/2 = x^2/(2*D).
- Hence k = -D/2 and a = 1/(2*D).
- By translation of vertex from (0,-D/2) to (h,k) we get the proof.
Go to Begin
AN 06 14. Example : Convert y = a*x^2 + b*x + c to polar form
- The vertex is at x = -b/(2*a).
- The distance between F and directrix is D = 1/(2*a).
- Translate vertex to (0,-D/2) so that the origin is at (0,0).
- Hence we have y + D/2 = x^2/(2*D).
- Add y^2 to both sides and we have
- y^2 + 2*D*y + D^2 = x^2 + y^2.
- Polar coordinates : x=r*cos(A) and y=R*sin(A).
- Hence (y+D)^2 = R^2.
- Take square root on both side we have y + D = R.
- or R*sin(A) + D = R.
- or R = D/(1-sin(A)).
- Since D = 1/(2*a), hence R = 1/(2*a*(1-sin(A)).
Other method
- Since origin at F(0,0) and the parabola opens upward.
- Hence R = D/(1-sin(A)).
- Since D=1/(2*a), hence R = 1/(2*a*(1-sin(A))
Go to Begin
AN 06 15. Example : Draw tangent to parabola using reflection
Figure 5
Tangent by reflection
- Draw a parabola using y-k = a*(x-h)^2.
- Draw a point P(x,y) on the parabola.
- Draw focus F at x = h and y = k+D/2 where D = 1/(2*a).
- Draw a line PF.
- Draw a line PQ parallel to the principal axis.
- Bisect angle FPQ.
- Draw line PR perpendicular to the bisector.
- By the law of reflection, This line is the requred tangent.
Go to Begin
Q16. Example : Convert R = D/(1-sin(A)) to y = F(x).
- The parabola opens upward and origin at F(0,0).
- Hence vertex is at (0,-D/2).
- Hence the equation is y + D/2 = x^2/(2*D).
Second method
- R*(1-sin(A)) = D.
- R - R*sin(A) = D. or R = (D+y).
- Squre both sides we get
- x^2 + y^2 = (D+y)^2.
- or x^2 + y^2 = D^2 + 2*D*y + y^2.
- or x^2 = 2*D*y + D^2.
- Hence y + D/2 = x^2/(2*D).
Go to Begin
AN 06 17. Elliminate x*y term in F(x,y)
Go to Begin
AN 06 18 Compare y + D/2 = x^2/(2*D) with R = D/(1-sin(A)).
Comparison
- Equation : y + D/2 = x^2/(2*D)
- Vertex is at (0,-D/2).
- It opens upward.
- Focus is at (0,0).
- Principal axis is x = 0.
- Equation of directrix is y = -D.
- Polar form : R = D/(1-sin(A))
- Focus is at (0,0).
- Equation of directrix is y = -D.
- It opens upward.
- Principal axis is x = 0.
Go to Begin
AN 06 19 Example : Stduy parabola of y = F(x) = x^2 - 6*x + 8.
[Questions]
- Find coordinates of vertex.
- Find the principal axis.
- Find distance from focus to directrix.
- Find the coordinates of focus.
- Find the equation of directrix.
- Find the openning.
- Find the Polar form.
[Answers]
- Vertex : xv=-(-6)/2 = 3 and yv = F(3) = -1.
- Principal axis : x = 3.
- Distance from focus to directrix : D = 1/(2*1) = 0.5.
- Coordinates of focus : xf = 3, yf = -1 + 0.25.
- Equation of directrix : y = -1 - 0.25 = -1.25.
- It opens upward.
- Polar form : R = 0.5/(1-sin(A))
Go to Begin
AN 06 20 Application of parabolic surface
- If light source is at focus, reflected light will be parallel to principal axis.
- Applications
- Headlight of vehicle.
- Search light.
Go to Begin
AN 06 21 Three points define a parabola.
- Substitute 3 points into y = a*x^2 + b*x + c
- We have 3 linear equations.
- Solve these 3 linear equations to get a,b and c.
- Computation tool : MD2002 program 03 11.
Go to Begin
Q22. Subjects of parabola on internet
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Subjects |
Definition and examples of circle on internet
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Subjects |
Graphs in Analytic geometry : Section 6
Go to Begin
AN 06 23. References
On PC computer
- Sketch R = D/(1-sin(A)) .................. MD2002 ZM40 08
- Conic Section ............................ MD2002 ZM06 and ZM40
- Elliminate x*y terms in F(x,y)=0 ......... MD2002 ZM34 12
- See keywords parabola in program index
- See contents of ZMxx in
Subjects and index on PC computer
Go to Begin
AN 06 00. Outline : Equation of parabola (x2,y2)
Terms about parabola
- Principal axis
- The curve of parabola is symmetrical to the principal axis
- The focus and vetex are on principal axis
- The directrix is perpendicular to the principal axis
- Focus
- It is on the principal axis
- It has distance from directrix equal D
- It has distance from vertex equal D/2
- Vertex
- It is on the principal axis
- It has distance from focus or directrix equal D/2
- It is between focus and directrix
- Directrix
- Directrix is perpendicualr to principal axis
- It has a distance D from the focus
- It has a distance D/2 from the vertex
- D = 1/(2*a) if y = a*x^2 + b*x + c
- Distance from focus to directrix is D = 1/(2*a)
- Locus : Point on curve has same distance from focus and directrix
Polynomial form
- y = a*x^2 + b*x + c
- y' = 2*a*x is the slope
Vertex form
- y - k = (x - h)^2/(2*D) where D is the distance from focus to directrix
- (h, k) is the vertex
- Compare with polynomial form we have D = 1/(2*a) or a = 1/(2*D)
- The focus is at x = h and y = k + D/2
- Equation of directrix is y = k - D/2
Polar form
- R = D/(1 - sin(A)). It opens upward
- R = D/(1 + sin(A)). It opens Downward
- R = D/(1 - cos(A)). It opens to right
- R = D/(1 + cos(A)). It opens to left
Special form : R = sec(A/2)^2
- Use sec(t) = 1/cos(t) and half angle formula cos(t/2) = Sqr((1 + cos(t))/2)
- Hence R = 2/(1 + cos(A))
- D = 2 and it open to left
Special form : R = csc(A/2)^2
- Use csc(t) = 1/sin(t) and half angle formula sin(t/2) = Sqr((1 - cos(t))/2)
- Hence R = 2/(1 - cos(A))
- D = 2 and it open to right
Formula
- If y-k = a*(x-h)^2. then
- Principal axis is x = h.
- Vertex is at (h,k).
- Focus F to directrix is D = 1/(2*a).
- Equation of directrix is y = k-D/2.
- Focus is at (h,k+D/2).
- Slope is y' = 2*a*(x-h)
- Polar form
- R = D/(1-sin(A)).
- R = D/(1+sin(A)).
- R = D/(1-cos(A)).
- R = D/(1+cos(A)).
Go to Begin
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