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Precalculus : Determine the equation of a parabola and graph a parabola

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Example Questions

Example Question #31 : Determine The Equation Of A Parabola And Graph A Parabola

Find the focus and the directrix of the following parabola: y^2=16x .

Possible Answers:

Focus: (0,-4)

Directrix: x=-4

Focus: (-4,0)

Directrix: x=4

Focus: (4, 0)
Directrix: x=-4

Focus: (0,4)

Directrix: x=4

Correct answer:

Focus: (4, 0)
Directrix: x=-4

Explanation:

To find the focus from the equation of a parabola, first set the equation to resemble the form y^2=4px where represents any numerical value.

For our problem, it is already in this form.

Therefore,

16=4p .

Solve for p: 16=4p then 4=p .

The focus for this parabola is given by (0,p) .

So, (0,4) is the focus of the parabola.

The directrix is represented as x= -p .

Therefore, the directrix for this problem is x=-4 .

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Example Question #32 : Determine The Equation Of A Parabola And Graph A Parabola

Find the directerix of the parabola with the following equation:

(x-5)^2=20y

Possible Answers:

x=-5

y=5

y=-5

x=5

Correct answer:

y=-5

Explanation:

Recall the standard form of the equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h, k) is the vertex of the parabola and |p| gives the focal length.

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

For the parabola in question, the vertex is (5, 0) and p=5 . This parabola will open up. Because the parabola will open up, the directerix will be located units down from the vertex. The equation for the directerix is then y=-5 .

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Example Question #31 : Determine The Equation Of A Parabola And Graph A Parabola

Find the directerix for the parabola with the following equation:

(x-5)^2=4(y+1)

Possible Answers:

x=-2

x=2

y=2

y=-2

Correct answer:

y=-2

Explanation:

Recall the standard form of the equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h, k) is the vertex of the parabola and |p| gives the focal length.

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

For the parabola in question, the vertex is (5, -1) and p=1 . This parabola will open up. Because the parabola will open up, the directerix will be located unit down from the vertex. The equation for the directerix is then y=-2 .

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Example Question #32 : Determine The Equation Of A Parabola And Graph A Parabola

Find the directerix of the parabola with the following equation:

x^2-4x+16y-76=0

Possible Answers:

x=9

x=1

y=1

y=9

Correct answer:

y=9

Explanation:

Recall the standard form of the equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h, k) is the vertex of the parabola and |p| gives the focal length.

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

Start by putting the equation in th estandard form of the equation of a vertical parabola.

Isolate the terms to one side.

x^2-4x+16y-76=0

x^2-4x=-16y+76

Complete the square for the terms. Remember to add the same amount on both sides!

x^2-4x+4=-16y+76+4

x^2-4x+4=-16y+80

Factor out both sides of the equation to get the standard form of a vertical parabola.

(x-2)^2=-16(y-5)

For the parabola in question, the vertex is (2, 5) and p=-4 . This parabola will open down. Because the parabola will open down, the directerix will be located units above the vertex. The equation for the directerix is then y=9 .

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Example Question #35 : Determine The Equation Of A Parabola And Graph A Parabola

Find the focus of the parabola with the following equation:

x^2+10x-4y+17=0

Possible Answers:

(5, 4)

(-5, -1)

(-5, 1)

(5, 2)

Correct answer:

(-5, -1)

Explanation:

Recall the standard form of the equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h, k) is the vertex of the parabola and |p| gives the focal length.

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

Start by putting the equation into the standard form.

Isolate the terms on one side.

x^2+10x-4y+17=0

x^2+10x=4y-17

Complete the square. Remember to add teh same amount on both sides!

x^2+10x+25=4y-17+25

x^2+10x+25=4y+8

Factor both sides of the equation to get the standard equation for the parabola.

(x+5)^2=4(y+2)

For the parabola in question, the vertex is (-5, -2) and p=1 . This parabola will open up. Because the parabola will open up, the focus will be located unit up from the vertex. The focus is then located at (-5, -1) .

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Example Question #35 : Determine The Equation Of A Parabola And Graph A Parabola

Find the focus of the parabola with the following equation:

(x-9)^2=-8(y-5)

Possible Answers:

(9, 3)

(9, -3)

(9, 11)

(9, 5)

Correct answer:

(9, 3)

Explanation:

Recall the standard form of the equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h, k) is the vertex of the parabola and |p| gives the focal length.

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

For the parabola in question, the vertex is (9, 5) and p=-2 . This parabola will open down. Because the parabola will open down, the focus will be located units down from the vertex. The focus is then located at (9, 3) .

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Example Question #51 : Parabolas

Find the focus of the parabola with the following equation:

(x+2)^2=-12y

Possible Answers:

(-2, -3)

(-2, 2)

(-2, 0)

(-2, -2)

Correct answer:

(-2, -3)

Explanation:

Recall the standard form of the equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h, k) is the vertex of the parabola and |p| gives the focal length.

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

For the parabola in question, the vertex is (-2, 0) and p=-3 . This parabola will open down. Because the parabola will open down, the focus will be located units down from the vertex. The focus is then located at (-2, -3)

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Example Question #52 : Parabolas

Find the directerix of the parabola with the following equation:

(y-3)^2=-4x

Possible Answers:

y=1

x=1

y=-1

x=-1

Correct answer:

x=1

Explanation:

Recall the standard form of the equation of a horizontal parabola:

(y-k)^2=4p(x-h) , where (h, k) is the vertex of the parabola and |p| is the focal length.

When p>0 , the parabola opens to the right.

When p<0 , the parabola opens to the left.

For the given parabola, the vertex is (0, 3) and p=-1 . This means the parabola is opening to the left and that the directerix will be located unit to the right of the vertex. The directerix is then x=1 .

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Example Question #53 : Parabolas

Find the directerix of the parabola with the following equation:

(y-4)^2=-8(x+1)

Possible Answers:

x=0

x=2

x=1

x=-1

Correct answer:

x=1

Explanation:

Recall the standard form of the equation of a horizontal parabola:

(y-k)^2=4p(x-h) , where (h, k) is the vertex of the parabola and |p| is the focal length.

When p>0 , the parabola opens to the right.

When p<0 , the parabola opens to the left.

For the given parabola, the vertex is (-1, 4) and p=-2 . This means the parabola is opening to the left and that the directerix will be located units to the right of the vertex. The directerix is then x=1 .

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Example Question #1841 : Pre Calculus

Find the directerix of the parabola with the following equation:

y^2+2y-12x+13=0

Possible Answers:

x=3

x=-1

x=4

x=-2

Correct answer:

x=-2

Explanation:

Recall the standard form of the equation of a horizontal parabola:

(y-k)^2=4p(x-h) , where (h, k) is the vertex of the parabola and |p| is the focal length.

When p>0 , the parabola opens to the right.

When p<0 , the parabola opens to the left.

Put the given equation into the standard form. Start by isolating the terms to one side.

y^2+2y-12x+13=0

y^2+2y=12x-13

Complete the square. Remember to add the same amount to both sides of the equation.

y^2+2y+1=12x-13+1

y^2+2y+1=12x-12

Factor both sides of the equation to get the standard form of the equation of a horizontal parabola.

(y+1)^2=12(x-1)

For the given parabola, the vertex is (1, -1) and p=3 . This means the parabola is opening to the right and that the directerix will be located units to the left of the vertex. The directerix is then x=-2 .

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Precalculus : Determine the equation of a parabola and graph a parabola

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #31 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #32 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #31 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #32 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #35 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #35 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #51 : Parabolas

Example Question #52 : Parabolas

Example Question #53 : Parabolas

Example Question #1841 : Pre Calculus

All Precalculus Resources

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