Years
>
Year 12>
Locus and The Parabola>
Graphing Parabolas (Focus and Directrix)>
Graphing Parabolas (Focus and Directrix) 1Graphing Parabolas (Focus and Directrix) 1
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 5 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- 5
- Answered
- Review
-
Question 1 of 5
1. Question
Which of the following will most likely be the graph of`x^2=4ay`Hint
Help VideoCorrect
Well Done!
Incorrect
In a quadratic equation, if the coefficient of `y` is positive, the parabola will be concave up. If it is negative, the parabola will be concave downNotice that there is no constant term (term without a variable).`x^2` `=` `4ay` This means that the vertex is at `(0,0)`, the originAlso, take note of the following components:Focus `(0,a)`this point lies in the parabola’s axis of symmetry and is at equal distance with all points on the parabolaDirectrix `y=-a`this line lies in the opposite direction of the parabola and is at equal distance with all points on the parabolaNext, check the sign of the coefficient of `y`.`x^2` `=` `4a``y` Hence, the parabola is concave up.Therefore, it would make sense to say that the graph for `x^2=4ay` is
-
Question 2 of 5
2. Question
Sketch the graph of `x^2=16y`Hint
Help VideoCorrect
Nice Job!
Incorrect
Standard Form (Concave Up)
`x^2=4``a``y`Focus
`(0,``a``)`Standard Form (Concave Down)
`x^2=-4``a``y`Focus
`(0,-``a``)`First, determine which standard form applies to the given equation.`x^2` `=` `16y` Since the coefficient of `y` is positive, use `x^2=4``a``y`This also means that the parabola will be concave upSolve for the focal length, `a`, by comparing the given equation to the standard form`x^2` `=` `4a``y` `x^2` `=` `16``y` `4a` `=` `16` `4a``divide4` `=` `16``divide4` Divide both sides by `4` `a` `=` `4` Identify the focus and directrix.Focus:`(0,``a``)` becomes `(0,``4``)`Directrix:`y=-``a` becomes `y=-``4`Start graphing the parabola by plotting the focus and directrix
Finally, draw a parabola that is concave up
-
Question 3 of 5
3. Question
Sketch the graph of `x^2=-8y`Hint
Help VideoCorrect
Excellent!
Incorrect
Standard Form (Concave Up)
`x^2=4``a``y`Focus
`(0,``a``)`Standard Form (Concave Down)
`x^2=-4``a``y`Focus
`(0,-``a``)`First, determine which standard form applies to the given equation.`x^2` `=` `-8y` Since the coefficient of `y` is negative, use `x^2=-4``a``y`This also means that the parabola will be concave downSolve for the focal length, `a`, by comparing the given equation to the standard form`x^2` `=` `-4a``y` `x^2` `=` `-8``y` `-4a` `=` `-8` `-4a``divide-4` `=` `-8``divide-4` Divide both sides by `-4` `a` `=` `2` Identify the focus and directrix.Focus:`(0,-``a``)` becomes `(0,-``2``)`Directrix:`y=``a` becomes `y=``2`Start graphing the parabola by plotting the focus and directrix<
Finally, draw a parabola that is concave down
-
Question 4 of 5
4. Question
Sketch the graph of `y^2=20x`Hint
Help VideoCorrect
Fantastic!
Incorrect
Standard Form (Concave Right)
`y^2=4``a``x`Focus
`(``a``,0)`Standard Form (Concave Left)
`y^2=-4``a``x`Focus
`(-``a``,0)`First, determine which standard form applies to the given equation.`y^2` `=` `20x` Since the coefficient of `x` is positive, use `y^2=4``a``x`This also means that the parabola will be concave rightSolve for the focal length, `a`, by comparing the given equation to the standard form`y^2` `=` `4a``x` `y^2` `=` `20``x` `4a` `=` `20` `4a``divide4` `=` `20``divide4` Divide both sides by `4` `a` `=` `5` Identify the focus and directrix.Focus:`(``a``,0)` becomes `(``5``,0)`Directrix:`x=-``a` becomes `x=-``5`Start graphing the parabola by plotting the focus and directrix
Finally, draw a parabola that is concave right
-
Question 5 of 5
5. Question
Sketch the graph of `y^2=-2x`Hint
Help VideoCorrect
Keep Going!
Incorrect
Standard Form (Concave Right)
`y^2=4``a``x`Focus
`(``a``,0)`Standard Form (Concave Left)
`y^2=-4``a``x`Focus
`(-``a``,0)`First, determine which standard form applies to the given equation.`y^2` `=` `-2x` Since the coefficient of `x` is negative, use `y^2=-4``a``x`This also means that the parabola will be concave leftSolve for the focal length, `a`, by comparing the given equation to the standard form`y^2` `=` `-4a``x` `y^2` `=` `-2``x` `-4a` `=` `-2` `-4a``divide-4` `=` `-2``divide-4` Divide both sides by `-4` `a` `=` `1/2` Identify the focus and directrix.Focus:`(-``a``,0)` becomes `(-``1/2``,0)`Directrix:`x=``a` becomes `x=``1/2`Start graphing the parabola by plotting the focus and directrix
Finally, draw a parabola that is concave left