Q1. Defintion of locus of a parabola
A1. 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.
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Q2. Equations of Parabola
A2. Answers
[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.
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Q3. 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.
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Q4. Equation in parametric form
A4. Answer : It is not often useful as ellipse
- x = F(t).
- y = G(t).
- y-k = a*(x-h)^2
- (h,k) is the vertex.
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Q5. Example : One 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.
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Q6. 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).
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Q07. Example : Sketch a parabola using ruler :
Figure 3
[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
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Q8. Example : Prove that R = csc(A/2)^2 is parabola
- 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.
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Q9. Example : Prove that R = sec(A/2)^2 is parabola
- 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.
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Q10. Example : If y = x^2 draw tangent to curve at x=3
- 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.
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Q11. 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)).
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Q12. References :
- 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
Index File
- See contents of ZMxx in
Contents by chapres
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Q13. 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.
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Q14. 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))
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Q15. 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.
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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).
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Q17. Elliminate x*y term in F(x,y)
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Q18 Compare y+D/2 = x^2/(2*D) with R = D/(1-sin(A)).
- 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 -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.
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Q19 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))
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Q20 Application of parabolic surface.
- If light source is at focus, reflected light will be parallel to principal axis.
- Applications
- Headlight of vehicle.
- Search light.
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Q21 Three points define a parabola.
- Substitute 3 points in to y = a*x^2 + b*x + c and get 3 linear equations.
- Solve these 3 linear equations to get a,b and c.
- Computation tool : MD2002 program 03 11.
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