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Precalculus : Conic Sections

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Example Questions

Example Question #42 : Determine The Equation Of A Parabola And Graph A Parabola

Find the focus of the parabola with the following equation:

(y-5)^2=20(x-7)

Possible Answers:

(12, 5)

(2, 4)

(-5, 5)

(2, 5)

Correct answer:

(12, 5)

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 (7, 5) and p=5 . This means the parabola is opening to the right and that the focus will be located units to the right of the vertex. The focus is then located at (12, 5) .

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Example Question #1843 : Pre Calculus

Find the focus of the parabola with the following equation:

y^2+12y-24x+60=0

Possible Answers:

(1, -11)

(-5, -6)

(7, -6)

(1, -1)

Correct answer:

(7, -6)

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.

Start by putting the equation into the standard form of the equation of a horizontal parabola.

Isolate the terms on one side.

y^2+12y-24x+60=0

y^2+12y=24x-60

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

y^2+12y+36=24x-60+36

y^2+12y+36=24x-24

Factor both sides of the equation to get the equation in the standard form.

(y+6)^2=24(x-1)

For the given parabola, the vertex is (1, -6) and p=6 . This means the parabola is opening to the right and that the focus will be located units to the right of the vertex. The focus is then located at (7, -6) .

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Example Question #1844 : Pre Calculus

Find the focus and directrix of the parabola

$y = \frac{(x-7)^2 }{4}$ .

Possible Answers:

Focus: (7,1) ; Directrix y = -1

Focus: (7,1) ; Directrix: y=1

Focus: (7,-1) ; Directrix: y = 1

Focus: (7,-4) ; Directrix: y = 4

Focus: (7, 4) ; Directrix: y = 4

Correct answer:

Focus: (7,1) ; Directrix y = -1

Explanation:

We can first re-write the equation by multiplying both sides by 4:

4y = (x-7)^2

This equation tells us that the vertex for the parabola is (7, 0)

We can also find the distance from the vertex to the focus and the directrix by setting up the equation 4 = 4p . This gives us 1 = p , so the focus and the directrix are both 1 away from the vertex.

Since this equation is positive, the parabola opens up. This means the directrix is one below the vertex, so it is at y = -1 . The focus is one above the vertex, so it is at (7,1) .

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Example Question #44 : Determine The Equation Of A Parabola And Graph A Parabola

Find the focus and directrix of the parabola y = x^2 + 4x +10 .

Possible Answers:

Focus: (-2, 5.75) ; Directrix: y = 6.25

Focus: (-2, 7) ; Directrix: y = 5

Focus: (-2, 6.25) ; Directrix: y = 5.75

Focus: (2, -5.75) ; Directrix: y = -6.25

Focus: (2, 7) ; Directrix: y = 5

Correct answer:

Focus: (-2, 6.25) ; Directrix: y = 5.75

Explanation:

Before we can start determining the focus and directrix, we need to re-write this equation in standard form. That way we can see the vertex and we can more easily determine the distance from the vertex and the focus/directrix.

To do this, we should complete the square. First subtract 10 from both sides:

y - 10 = x^2 + 4 x

Now we need to figure out what to add to both sides in order to make the right side a perfect square. We should add 4, since half of the 4 in 4x is 2, and 2^2 = 4

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

We can simplify the left side and re-write the right side as a binomial squared:

y - 6 = (x+2)^2

Interpreting this, we determine that the vertex is (-2, 6) . On the left side, y-6 is multiplied by 1, so 4p = 1 where p measures the distance from the vertex to the focus/directrix. Solving this gives us $p = \frac{1}{4}= 0.25$ .

Since this is a positive, upward-opening parabola, the directrix is below the vertex and the focus is above. The focus is then located at (-2, 6+0.25 ) = (-2, 6.25) . The directrix is then located at y = 6 - 0.25 = 5.75 .

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Example Question #41 : Determine The Equation Of A Parabola And Graph A Parabola

Determine the direction in which the following parabola opens.

y=-5x^2

Possible Answers:

Leftwards

Upwards

Rightwards

Downwards

Correct answer:

Downwards

Explanation:

For the function

y=ax^2

The parabola opens upwards if a>0

and downards for a<0

Because

-5<0

The parabola opens downwards.

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Precalculus : Conic Sections

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #42 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #1843 : Pre Calculus

Example Question #1844 : Pre Calculus

Example Question #44 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #41 : Determine The Equation Of A Parabola And Graph A Parabola

All Precalculus Resources

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