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PARABOLA
- A parabola is a set of all points in the xy- plane that is equidistant from a fixed point and a fixed line. The fixed point is called the focus and the fixed distance is called the directrix.
- The closest point from the focus to the parabola is called the vertex.
- The distance between the focus and the parabola is called the focal distance denoted by c. ๏ Formulas to remember: Exercises 1. A. Graph the parabola. Determine the foci, vertex, and directrix.
1. x^2 = 10 y
Sol:
2. x^2 =โ 12 y
Sol:
3. 2 y^2 = 16 x
Sol:
4. y^2 =โ 14 x
Sol:
5. ( x + 3 )^2 = 4 ( y โ 1 )
Sol:
Sol:
Sol: Vertical parabolas:
( x โ h )
2
= 4 c ( y โ k )
- f ( h , k + c )
- d : y = k โ c Horizontal parabolas:
( y โ k )
2
= 4 c ( x โ h )
- f ( h + c , k )
- d : x = h โ c To find x-intercept, set y to 0. To find y-intercept, set x to 0.
Sol:
Sol:
Sol:
Sol:
Sol:
Vertex (-1, -4), focus 3 units to the left of the vertex Sol: ๏ In a parabola that opens to the left, the focus is always left of the vertex and the directrix is always right of the vertex, and as such, denotes a negative c.
Focus (1,3), directrix 6 units to the left of the focus Sol: ๏ In a parabola that opens to the right, the focus is always right of the vertex and the directrix is always left of the vertex, and as such, denotes a positive c.
Focus (2,5), directrix 6 units to the right of the vertex Sol:
22. Directrix y =โ 5 , vertex 2 units above the
directrix Sol:
23. Directrix x = 5 , vertex 3 units to the left of
the focus Sol:
- Vertex (-2, -4), vertical axis, through the point (6, 4/3) Sol: ๏ The axis of symmetry (simply axis) of a parabola is the line that equally divides the parabola at its vertex. ๏ A vertical axis denotes a vertical parabola. ๏ โThrough the pointโ usually denotes substituting the x and y coordinates in an equation.
- Vertex (3,2), vertical directrix, through the point (-3/2,-4) Sol: ๏ A vertical directrix denotes a horizontal parabola.
- Vertex (-1,-6), through the points (-5,-4) and (11,12) Sol: ๏ In problems where there is no indicator telling whether the parabola is either vertical or horizontal, sketch the graph and analyze the relationship of the points and the vertex. The vertex should always be the lowest or highest point in a vertical parabola and the leftmost or rightmost point in a horizontal parabola. ๏ Here, we see that the graph is a vertical parabola that opens upward.
- A parabola which opens upward has its
vertex on the line y = x โ 2 and focus on
the line y = 9 โ 2 x.
Sol:
- Find the equation of all parabolas having a vertex (-2,-3), passing through (2,1), and whose directrix is horizontal or vertical. Sol: ๏ There are two solutions and answers to this problem. VERTICAL:
๏ The common axis is y =โ 3 , which also
doubles as location of the circleโs center. To find the center, get the midpoint of the two points. C. Solve the following problems.
- The bulb of a flashlight is placed at the focus above the parabolic reflector. If the surface is 5 inches across and 1.5 deep at the vertex, how far is the bulb above the vertex? Sol: ๏ In solving word problems like these, the first step is to always identify the given variables and what they represent in our equation. Always pay attention to the unit used. ๏ In the case wherein the problem is not given coordinates or is simple enough not to use them, the std. eqโn. for parabolas can be simplified to. ๏ In hindsight, โ5 inches acrossโ might sound like the true value of x. The value of x is the number of units a point is away from its respective axis. Since parabolas are divided by their axes, we have to divide 5 by 2, hence.
- The bulb of a flashlight is placed at the focus above the parabolic reflector. If the surface is 4.5 inches across and the bulb is 1 inch above the vertex, how deep is the reflector at the vertex? Sol:
- A satellite dish shaped like a paraboloid has a width of 9 ft across and a depth of 2 ft above the vertex. If the receiver is placed at the focus, how high above the vertex is the receiver? Sol:
- A satellite dish shaped like a paraboloid has a width of 8ft across with the receiver (at the focus) 2.5 ft above the vertex. How deep is the dish above the vertex? Sol:
- Two towers of a suspension bridge are 1000 m apart and are 160 m high. The cable between the two towers hangs in the shape of a parabola, which, at its lowest, is 40 m above the road. Sol: ๏ To find the real y-component of the parabola, you need to properly illustrate which heights are which. If lowest point of the parabola is 40m above the road and its highest peaking at 160m, then by subtracting the lowest from the highest,
y = 160 โ 40 , we then get y =120.
๏ The lowest or highest point is the vertex,
therefore making our vertex V (0,40).
๏ With our x at 500 units away from the y- axis ( ), and our maximum height at
y = 160 , we get two points in our parabola,
namely A (โ500,160) and B (500,160).
a. Find the equation of this parabola, if we put a coordinate system so that the axis of the parabola is the y-axis, and the road is on the x- axis. Sol:
b. How high is the cable 100 m away from a tower? Sol:
- Two towers of a suspension bridge are 800 m apart and are 180 m high. The cable between the two towers hangs in the shape of a parabola, which, at its lowest, just touches the road. How high above the road is the cable 300 m away from the center? Sol:
- A ball was thrown upward from ground level. As it rose and eventually fell back to the ground, its path traced a parabola. Suppose it reached a maximum elevation of 12 ft and it landed 20 ft away from where it was thrown up. Sol: ๏ If you encounter cases where the axis of symmetry is no longer the y-axis, the axis
is still the line that divides x. If x = 20 , then
the axis is at (10,0) with the true value of
x at x = 10.
a. Find the equation of the parabola, if a coordinate system super- imposed, with the ball starting at the origin and the x-axis at ground level. Sol: b. What was the elevation of the ball 4 ft away (horizontally) from where it was thrown? Sol:
- A ball was thrown upward from ground level and its path has the shape of a parabola. If it went as high as 15 ft and then fall back to the ground 24 ft away (horizontally) from where it was thrown), find the equation of the parabola. Assume the x-axis is at the ground and the ball was thrown from the origin.
Sol: b. Equate the two expressions in (a). Manipulate this equation to derive the standard equation of the parabola. Sol: c. Would your computations change if c < 0? Sol: Yes, it will change
42. The segment through the focus of a parabola which is parallel to the directrix (hence, perpendicular to the axis) is called the latus rectum. Sol: ๏ To find the latus rectum of a parabola, one needs to find the focus first. Then, draw a line parallel to the directrix. The points where the line intercepts the parabola are the endpoints of the latus rectum. a. Find the endpoints of the latus rectum of the parabola . How long is the latus rectum? Sol: ๏ง Latus rectum, should we
denote as simply l is located
at the line y =3. Find the
endpoints by substituting y to 3. Then, find its length by getting the distance of the two points. b. How long is the latus rectum of the parabola? Sol: This pdf is not free from errors. Please correct on your own if you see them or inform me if you want a new copy. I tried my best in proofreading this document but I may have missed some things. Please do not redistribute or share. If someone you know wants a copy, refer them to me. Thanks!
- Bernice Danielle E. Castillo 12STEM- Extra paper for scratch solutions
