Parabola Directrix Calculator

Enter the parabola equation to calculate its directrix equation and see the step-by-step resolution along with the graph.

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Solved Exercises

Determine the directrix equation for the parabola $ y=x^2+6x+5. $

Result

The directrix equation of the given parabola is:

$$ y = -\dfrac{17}{4} = -4.25 $$

Step-by-step resolution

1. Obtain the vertex coordinates.

We will start by examining the entered equation:

$$ y = x^2 + 6x + 5 $$

Since the equation is in standard form $ y = ax^2 + bx + c $, we will extract the leading coefficient which will be useful later:

$$ a = 1 $$

We determined that the vertex of the parabola is $ V\left(-3, -4\right) $. If you want to see how it was calculated, you can check the vertex calculator.

2. Calculate the focal parameter (p).

The parameter p defines the distance from the vertex to the focus and the directrix. We find it using the coefficient 'a' through the following formula:

$$ p = \dfrac{1}{4a} \\[1em] p = \dfrac{1}{4(1)} \\[1em] p = \dfrac{1}{4} $$

3. Formulate the directrix equation.

Since the parabola is vertical (the x variable is squared), the directrix will be a horizontal line. Its equation is defined by subtracting the parameter p from the y-coordinate of the vertex (k):

$$ y = k - p \\[1em] y = -4 - \dfrac{1}{4} \\[1em] y = -\dfrac{17}{4} = -4.25 $$

Graph on the Cartesian plane of a vertical parabola opening upward and its directrix. — Graph of the parabola with the directrix

Calculate the directrix of the parabola $ y=-3x^2+6x-2. $

Answer

The directrix equation of the given parabola is:

$$ y = \dfrac{13}{12} \approx 1.08 $$

Step-by-step resolution

1. Obtain the vertex coordinates.

We will start by examining the entered equation:

$$ y = -3x^2 + 6x - 2 $$

Since the equation is in standard form $ y = ax^2 + bx + c $, we will extract the leading coefficient which will be useful later:

$$ a = -3 $$

We determined that the vertex of the parabola is $ V\left(1, 1\right) $.

2. Calculate the focal parameter (p).

The parameter p defines the distance from the vertex to the focus and the directrix. We find it using the coefficient 'a' through the following formula:

$$ p = \dfrac{1}{4a} \\[1em] p = \dfrac{1}{4(-3)} \\[1em] p = -\dfrac{1}{12} $$

3. Formulate the directrix equation.

Since the parabola is vertical (the x variable is squared), the directrix will be a horizontal line. Its equation is defined by subtracting the parameter p from the y-coordinate of the vertex (k):

$$ y = k - p \\[1em] y = 1 - \left(-\dfrac{1}{12}\right) \\[1em] y = \dfrac{13}{12} \approx 1.08 $$

Graph on the Cartesian plane of a vertical parabola opening downward and its directrix. — Graph of the parabola with its directrix

Obtain the directrix of the parabola knowing its vertex form equation $ (y+2)^2=8(x-5). $

Solution

The directrix equation of the given parabola is:

$$ x = 3 $$

Step-by-step resolution

1. Obtain the vertex coordinates.

We will start by examining the entered equation:

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

The equation is already in vertex form $ (y - k)^2 = 4p(x - h) $, which makes it easier to directly extract the key parameters. The vertex (h, k) of the parabola is located at:

$$ V\left(5, -2\right) $$

2. Calculate the focal parameter (p).

In the vertex form, the coefficient accompanying the linear term is equal to 4p. We equate this value to solve for the unknown p:

$$ 4p = 8 \\[1em] p = \dfrac{8}{4} \\[1em] p = 2 $$

3. Formulate the directrix equation.

Since the parabola is horizontal (the y variable is squared), the directrix will be a vertical line. Its equation is defined by subtracting the parameter p from the x-coordinate of the vertex (h):

$$ x = h - p \\[1em] x = 5 - 2 \\[1em] x = 3 $$

Graph on the Cartesian plane of a horizontal parabola opening to the right and its directrix. — Graph of the parabola with the directrix

Find the directrix equation of the parabola $ x=-\frac14y^2-2y+1. $

Result

The directrix equation of the given parabola is:

$$ x = 6 $$

Step-by-step resolution

1. Obtain the vertex coordinates.

We will start by examining the entered equation:

$$ x = -\dfrac{1}{4}y^2 - 2y + 1 $$

Since the equation is in standard form $ x = ay^2 + by + c $, we will extract the leading coefficient which will be useful later:

$$ a = -\dfrac{1}{4} $$

We determined that the vertex of the parabola is $ V\left(5, -4\right) $.

2. Calculate the focal parameter (p).

The parameter p defines the distance from the vertex to the focus and the directrix. We find it using the coefficient 'a' through the following formula:

$$ p = \dfrac{1}{4a} \\[1em] p = \dfrac{1}{4\left(-\dfrac{1}{4}\right)} \\[1em] p = -1 $$

3. Formulate the directrix equation.

Since the parabola is horizontal (the y variable is squared), the directrix will be a vertical line. Its equation is defined by subtracting the parameter p from the x-coordinate of the vertex (h):

$$ x = h - p \\[1em] x = 5 - \left(-1\right) \\[1em] x = 6 $$

Graph on the Cartesian plane of a horizontal parabola opening to the left and its directrix. — Graph of the parabola and the directrix

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Professor of Mathematics, graduated from the Faculty of Exact, Chemical and Natural Sciences of the National University of Misiones (UNAM). Developer and creator of RigelUp, dedicated to building tools for mathematical learning.