Parabola Focus Calculator

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

Quick Examples

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Solved Exercises

Calculate the focus of the parabola from its equation: $ y=x^2-4x+7. $

Result

The focus of the given parabola is:

$$ F\left(2, \dfrac{13}{4}\right) = \left(2, 3.25\right) $$

Step-by-step resolution

1. Identify the equation and the vertex.

The entered equation is:

$$ y = x^2 - 4x + 7 $$

The equation is in its standard form: $ y = ax^2 + bx + c $. Therefore, we extract the leading coefficient, as we will use it in the next step:

$$ a = 1 $$

The vertex of the parabola is located at $ V\left(2, 3\right) $. To see how to find it step by step, check the vertex calculator.

2. Find the focal parameter (p).

From coefficient a, we calculate the parameter p using this formula:

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

3. Calculate the focus coordinates.

Since it is a vertical parabola (the x variable is squared), to find the focus we must add the parameter p (with its sign) to the y-coordinate of the vertex:

$$ F(h, k + p) \\[1em] F\left(2, 3 + \dfrac{1}{4}\right) \\[1em] F\left(2, \dfrac{13}{4}\right) = \left(2, 3.25\right) $$

Graph on the Cartesian plane of a vertical parabola opening upward with equation, focus, and vertex. — Graph of the parabola and its focus

Determine the coordinates of the focus of the parabola whose equation is $ y=-2x^2-12x-14. $

Solution

The focus of the given parabola is:

$$ F\left(-3, \dfrac{31}{8}\right) = \left(-3, 3.875\right) $$

Step-by-step resolution

1. Identify the equation and the vertex.

The entered equation is:

$$ y = -2x^2 - 12x - 14 $$

The equation is in its standard form (polynomial): $ y = ax^2 + bx + c $. Therefore, we extract the leading coefficient, as we will use it in the next step:

$$ a = -2 $$

The vertex of the parabola is located at $ V\left(-3, 4\right) $.

2. Find the focal parameter (p).

From coefficient a, we calculate the parameter p using this formula:

$$ p = \dfrac{1}{4a} \\[1em] p = \dfrac{1}{4(-2)} \\[1em] p = -\dfrac{1}{8} $$

3. Calculate the focus coordinates.

Since it is a vertical parabola (the x variable is squared), to find the focus we must add the parameter p (with its sign) to the y-coordinate of the vertex:

$$ F(h, k + p) \\[1em] F\left(-3, 4 + \left(-\dfrac{1}{8}\right)\right) \\[1em] F\left(-3, \dfrac{31}{8}\right) = \left(-3, 3.875\right) $$

Graph on the Cartesian plane of a vertical parabola opening downward with equation, focus, and vertex. — Graph of the parabola and its focus

Find the focus of the parabola defined by the vertex form equation $ (y-3)^2=10(x+4). $

Answer

The focus of the given parabola is:

$$ F\left(-\dfrac{3}{2}, 3\right) = \left(-1.5, 3\right) $$

Step-by-step resolution

1. Identify the equation and the vertex.

The entered equation is:

$$ (y - 3)^2 = 10(x + 4) $$

The equation is in its vertex form: $ (y - k)^2 = 4p(x - h) $, where the vertex is (h, k). This structure allows us to extract the necessary parameters directly. In this case, the vertex is:

$$ V\left(-4, 3\right) $$

2. Find the focal parameter (p).

The coefficient multiplying the linear term in the original equation corresponds to 4p. We set it equal while keeping its sign to solve for p:

$$ 4p = 10 \\[1em] p = \dfrac{10}{4} \\[1em] p = \dfrac{5}{2} $$

3. Calculate the focus coordinates.

Since it is a horizontal parabola (the y variable is squared), to find the focus we must add the parameter p (with its sign) to the x-coordinate of the vertex:

$$ F(h + p, k) \\[1em] F\left(-4 + \dfrac{5}{2}, 3\right) \\[1em] F\left(-\dfrac{3}{2}, 3\right) = \left(-1.5, 3\right) $$

Graph on the Cartesian plane of a horizontal parabola opening to the right with equation, focus, and vertex. — Graph of the parabola with its focus

Calculate the position of the focus for the parabola described by the equation $ (x+2)^2=6(y-1). $

Result

The focus of the given parabola is:

$$ F\left(-2, \dfrac{5}{2}\right) = \left(-2, 2.5\right) $$

Step-by-step resolution

1. Identify the equation and the vertex.

The entered equation is:

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

The equation is in its vertex form: $ (x - h)^2 = 4p(y - k) $, where the vertex is (h, k). This structure allows us to extract the necessary parameters directly. In this case, the vertex is:

$$ V\left(-2, 1\right) $$

2. Find the focal parameter (p).

The coefficient multiplying the linear term in the original equation corresponds to 4p. We set it equal while keeping its sign to solve for p:

$$ 4p = 6 \\[1em] p = \dfrac{6}{4} \\[1em] p = \dfrac{3}{2} $$

3. Calculate the focus coordinates.

Since it is a vertical parabola (the x variable is squared), to find the focus we must add the parameter p (with its sign) to the y-coordinate of the vertex:

$$ F(h, k + p) \\[1em] F\left(-2, 1 + \dfrac{3}{2}\right) \\[1em] F\left(-2, \dfrac{5}{2}\right) = \left(-2, 2.5\right) $$

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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.