Exercise 1

Determine the equations of the following parabolas and indicate the values of their focal parameter, focus and directrix.

1 6y^2 - 12x = 0

2y^2 = -7x

 

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Exercise 2

Determine the equations of the parabolas using the information given:

1 The directrix is x = −3 and the focus is (3, 0).

2 The directrix is y = 4 and the vertex is (0, 0).

3 The directrix is y = −5 and the focus is (0, 5).

4 The directrix is x = 2 and the focus is (−2, 0).

5 The focus is (2, 0) and the vertex is (0, 0).

6 The focus is (3, 2) and the vertex is (5, 2).

7 The focus is (−2, 5) and the vertex is (−2, 2).

8 The focus is (3, 4) and the vertex is (1, 4).

Exercise 3

Calculate the vertex, focus and directrix of the following parabolas:

y^2 - 6y - 8x + 17 = 0

x^2 - 2x - 6y - 5 = 0

y = x^2 - 6x + 11

Exercise 4

Find the equation of the vertical parabola that passes through the points: A = (6, 1), B = (−2, 3) and C = (16, 6).

Exercise 5

Determine the equation of the parabola with a directrix of y = 0 and a focus at (2, 4).

Exercise 6

Determine the point(s) of intersection between the line r ≡ x + y − 5 = 0 and the parabola y² = 16x.

Exercise 7

Find the equation of the horizontal parabola that passes through the point (3, 4) and has its vertex at (0, 0).

Exercise 8

Determine the equation of the parabola with an axis parallel to the y-axis, vertex on the x-axis and which passes through the points A = (2, 3) and B = (−1, 12).

Exercise 9

Determine the equation of the parabola with a directrix of x + y − 6 = 0 and a focus at (0, 0).

 

Solution of exercise 1

Determine the equations of the following parabolas and indicate the values of their focal parameter, focus and directrix.

6y^2 - 12x = 0

Calculate the equations of the following parabolas

6y^2 = 12x               y^2 = 2x

2p = 2           p = 1

F = (\frac {1}{2} , 0)                  x = -\frac {1}{2}

2 2y^2 = -7x

Determine the equations of the following parabolas

2y^2 = -7x             y^2 = -\frac{7}{2}x

2p = \frac {7}{2}                 p = \frac{7}{4}

F = (-\frac{7}{8}, 0)             x = \frac {7}{8}

3   15x^2 = -42y

Determine the equations of the following parabolas

15x^2 = -42x          x^2 = -\frac {14}{5} y

2p = \frac {14}{5}           p = \frac {7}{5}

F = (0, - \frac {7} {10})              x = \frac {7}{10}

Solution of exercise 2

Determine the equations of the parabolas using the information given:

1 The directrix is x = −3 and the focus is (3, 0).

parabolas

p = d(F, r) = 6

y ^2 = 12x

2 The directrix is y = 4 and the vertex is (0, 0).

Equations of parabolas

\frac {p}{2} = 4

x^2 = -16y

3 The directrix is y = −5 and the focus is (0, 5).

Calculate the equations of parabolas

\frac {p}{2} = 5

x^2 = 20y

4 The directrix is x = 2 and the focus is (−2, 0).

Work out the equations of parabolas

p = 4

y^2 = -8x

5 The focus is (2, 0) and the vertex is (0, 0).

Solving parabolas

p = 4

y^2 = 8x

6 The focus is (3, 2) and the vertex is (5, 2).

Solving the equations of parabolas

\frac{p}{2} = 2

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

7 The focus is (−2, 5) and the vertex is (−2, 2).

Solving parabola equations

\frac {p}{2} = 3

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

8 The focus is (3, 4) and the vertex is (1, 4).

Solving parabolas

\frac{p}{2} = 2

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

 

Solution of exercise 3

Calculate the vertex, focus and directrix of the following parabolas:

y^2 - 6y - 8x + 17 = 0

Solving parabolas equations

(y^2 - 6y + 9) - 9 - 8x + 17 = 0

(y^2 - 6y + 9) = 8x - 8

(y - 3)^2 = 8 (x - 1)           V(1, 3)

2p = 8         \frac {p}{2} = 2

F = (1 + 2, 3)           F = (3, 3)

d = x = 1 - 2          d = x = -1

 

x^2 - 2x - 6y - 5 = 0

How to solve parabolas

(x^2 - 2x + 1) - 1 - 6y - 5 = 0

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

(x - 1)^2 = 6 (y + 1)       V(1, -1)

2p = 6             \frac{p}{2} = \frac{3}{2}

F = (1, -1 + \frac{3}{2})                  F = (1, \frac{1}{2})

d = y = -1 - \frac {3}{2}                  d = y = -\frac {5}{2}

y = x^2 - 6x + 11

How to solve parabolas equations

y = (x^2 - 6x + 9) - 9 + 11

y = (x^2 - 6x + 9) = y - 2

(x - 3)^2 = 1 \cdot (y - 2)        V(3, 2)

2p = 1             \frac{p}{2} = \frac {1}{4}

F (3, 2 + \frac {1}{4})             F(3, \frac{9}{4})

d = y = 2 - \frac {1}{4}              d = y = \frac {7}{4}

Solution of exercise 4

Find the equation of the vertical parabola that passes through the points: A = (6, 1), B = (−2, 3) and C = (16, 6).

y = ax^2 + bx + c

1 = a \cdot 36 + b \cdot 6 + c

1 = a \cdot 4 + b \cdot (-2) + c

6 = a \cdot 256 + b \cdot 16 + c

a = \frac {1}{24}       b = - \frac {10}{24}    c = 2

y = \frac{1}{24} x^2 - \frac {10}{24}x + 2

 

Solution of exercise 5

Determine the equation of the parabola with a directrix of y = 0 and a focus at (2, 4).

d(P, F) = d(P, d)

\sqrt{(x - 2)^2 + (y - 4)^2} = \frac {y} {\sqrt{0^2 + 1^2}}

y^2 = (x - 2)^2 + y^2 -8y + 16

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

Solution of exercise 6

Determine the point(s) of intersection between the line r ≡ x + y − 5 = 0 and the parabola y² = 16x.

solution

\begin{cases} y^2 = 16x \\ y = 5 - x\end{cases}          (5 - x)^2 = 16x

x^2 - 26x + 25 = 0

x_1 = 25      x_2 = 1

A = (25, 20)              B = (1, 4)

Solution of exercise 7

Find the equation of the horizontal parabola that passes through the point (3, 4) and has its vertex at (0, 0).

y^2 = 2px

4^2 = 2p \cdot 3

2p = \frac {16}{3}

y^2 = \frac {16}{3}x

Solution of exercise 8

Determine the equation of the parabola with an axis parallel to the y-axis, vertex on the x-axis and which passes through the points A = (2, 3) and B = (−1, 12).

Axis parallel to the y-axis       (x - a)^2 = 2p(y - b)

Vertex on the x-axis                    (a, 0)

\begin{cases}(2 - a)^2 = 2p (3 - 0) \\ (-1 - a)^2 = 2p (12 - 0) \end{cases}          \begin{cases}(4 + a)^2 - 4a= 6p  \\ 1 + a^2 + 2a = 24p \end{cases}

a_1 = 5            p_1 = \frac{3}{2}             a_2 = 1             p_2 = \frac {1}{6}

(x - 5)^2 = 3y              (x - 1)^2 = \frac {1}{3}y

Solution of exercise 9

Determine the equation of the parabola with a directrix of x + y − 6 = 0 and a focus at (0, 0).

d(P,F) = d(P,d)

\sqrt{(x - 0)^2 + y - 0)^2} = \frac {x +y - 6} {\sqrt{1^2 + 1^2}}

x^2 + y^2 - 2xy + 12x + 12y - 36 = 0

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.