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# Math1.2

## by wraithxjmin on Apr 13, 2008

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## Math1.2Presentation Transcript

• 1.3 : Parabolas
• If a plane intersects the cone when it is slanted the same as the side of the cone, (formally, when it is parallel to the slant height), the conic section is a parabola. This is shown below:
• Parabolas Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line called the directrix . Directrix The light source is the Focus The cross section of a headlight is an example of a parabola...
• Here are some other applications of the parabola...
• Directrix Focus d 1 d 1 d 2 d 2 d 3 d 3 Also, notice that the distance from the focus to any point on the parabola is equal to the distance from that point to the directrix... We can determine the coordinates of the focus, and the equation of the directrix, given the equation of the parabola.... Vertex Notice that the vertex is located at the midpoint between the focus and the directrix...
• Standard Equation of a Parabola: (Vertex at the origin) Equation Focus Directrix x 2 = 4py (0, p) y = –p Equation Focus Directrix y 2 = 4px (p, 0) x = –p (If the x term is squared, the parabola is up or down) (If the y term is squared, the parabola is left or right)
• Examples: Determine the focus and directrix of the parabola y = 4x 2 : Since x is squared, the parabola goes up or down… Solve for x 2 x 2 = 4py y = 4x 2 4 4 x 2 = 1/4y Solve for p 4p = 1/4 p = 1/16 Focus: (0, p) Directrix: y = –p Focus: (0, 1/16) Directrix: y = –1/16 Let’s see what this parabola looks like...
• Examples: Determine the focus and directrix of the parabola – 3y 2 – 12x = 0 : Since y is squared, the parabola goes left or right… Solve for y 2 y 2 = 4px – 3y 2 = 12x – 3y 2 = 12x –3 –3 y 2 = –4x Solve for p 4p = –4 p = –1 Focus: (p, 0) Directrix: x = –p Focus: (–1, 0) Directrix: x = 1 Let’s see what this parabola looks like...
• Examples: Write the standard form of the equation of the parabola with focus at (0, 3) and vertex at the origin. Since the focus is on the y axis,(and vertex at the origin) the parabola goes up or down… x 2 = 4py Since p = 3 , the standard form of the equation is x 2 = 12y Example: Write the standard form of the equation of the parabola with directrix x = –1 and vertex at the origin. Since the directrix is parallel to the y axis,(and vertex at the origin) the parabola goes left or right… y 2 = 4px Since p = 1 , the standard form of the equation is y 2 = 4x
• Equation Parabola at (h,k) If the vertex of the parabola is at (h,k), the standard equation for the parabola are as summarised below.
• Opens downwards y=k+p (h,k-p) (x-h) 2 =-4p(y-k) Opens upwards y=k-p (h,k+p) (x-h) 2 =4p(y-k) Opens to the left x=h+p (h-p,k) (y-k) 2 =-4p(x-h) Opens to the right x=h-p (h+p,k) (y-k) 2 =4p(x-h) Shape Directrix Focus Parabola
• Example 1
• State the vertex, focus and directrix of each of
Solution ( y−2) 2 =12( x−3) Vertex, V(h, k) = V(3, 2) P= 3, so focus, F( (h+p, 2) P(3+3, 2) P(6, 2 )
• Example 4
• Sketch the graph of x 2 + 8x + 4y + 12 = 0.
• Solution
• Write the equation in standard form.
• x 2 + 8x + 4y + 12 = 0.
• x 2 + 8x + 16 + 4y + 12 – 16 = 0.
• (x + 4) 2 + 4(y – 1) = 0
• (x + 4) 2 = – 4(y – 1)
• This is a parabolic equation with vertex
• V(– 4, 1), p = – 1 and F(– 4, 0).
• When y = 0, x = – 2 or x = – 6. The sketch of the graph is as follow,
• Symmetrical axis directrix V (–4, 1) (–6, 0) F (–4, 0)
• (–2, 0)
0
• Example 5
• Find the equation of a parabola which satisfies the following conditions, vertex , vertical symmetric axis and the parabola passes through point (3,6)
• Solution
• Standard equation of the parabola is
• Vertical symmetric axis
• The parabola passes through (3,6)
• The equation of the parabola is:
• Example 6 A necklace hanging between two fixed points A and B at the same level. The length of the necklace between the two point is 100 cm. The mid point of the necklaceis 8 cm below A and B. Assume that the necklace hangs in the form of parabolic curve, find the equation of the curve.
• Solution: 8cm 50cm 50cm A B y x