Question

Instructions: What is the equation for the parabola with the given focus and directrix? Focus: (-1/52,0) Directrix: x=1/52.

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Answer 1

The equation for the parabola with the given focus (-1/52, 0) and directrix x = 1/52 is y = (1/13)x

Explanation:

To find the equation for the parabola with the given focus and directrix, we can use the geometric definition of a parabola.

1. The focus of the parabola is (-1/52, 0).

2. The directrix of the parabola is x = 1/52.

3. The distance between the focus and the directrix is the same for any point on the parabola.

4. This distance is equal to the perpendicular distance from any point on the parabola to the directrix.

5. The vertex of the parabola is the midpoint between the focus and the directrix.

Vertex = (midpoint of focus and directrix) = ((-1/52 + 1/52)/2, (0 + 0)/2) = (0, 0)

6. The distance from the vertex to the focus is equal to the distance from the vertex to the directrix.

7. Therefore, the vertex is equidistant from the focus and the directrix, which means the parabola opens either upward or downward.

8. Since the directrix is a vertical line, the parabola opens either upward or downward along the y-axis.

9. Given that the focus is below the vertex and the directrix is above the vertex, the parabola opens upward.

The equation for a parabola with vertex (h, k) that opens upward can be written as (y - k) = 4a(x - h), where a is the distance from the vertex to the focus.

10. The vertex is (0, 0), and the focus is (-1/52, 0).

11. The distance from the vertex to the focus is a = 1/52.

12. Substituting the values into the equation, we have (y - 0) = 4(1/52)(x - 0).

13. Simplifying, the equation becomes y = (1/13)x.

Therefore, the equation for the parabola with the given focus (-1/52, 0) and directrix x = 1/52 is y = (1/13)x.

Answer 2

`y^2&=-(1)/(13)x`

Explanation:

Given values:

`\bullet\;\;\textsf{Focus:}\quad \left(-(1)/(52),0\right)`

`\bullet\;\;\textsf{Directrix:} \quad x=(1)/(52)`

The directrix of a parabola is a fixed line outside of the parabola that is perpendicular to the axis of symmetry. Therefore, as the directrix is vertical (x = 1/52), the parabola has a horizontal axis of symmetry. This means that the parabola is horizontal (opens left or right).

The focus of a parabola is a fixed point located on the axis of symmetry, such that all points on the parabola are equidistant to both the focus and the directrix. It is located inside the parabola. Therefore, as the focus is to the left of the directrix, the parabola opens to the left. The axis of symmetry of the given parabola is the y-value of the focus, y = 0.

The vertex of a sideways parabola is its turning point. As the parabola opens to the left, its vertex is the parabola's rightmost point. The vertex is located on the axis of symmetry. Therefore, the y-coordinate of the vertex of the given parabola is y = 0.

The standard form of a parabola with a horizontal axis of symmetry is:

`\boxed{(y-k)^2=4p(x-h)}`

where:

• p ≠ 0
• Vertex = (h, k)
• Focus = (h+p, k)
• Directrix: x = (h - p)
• Axis of symmetry: y = k

As the focus is (-1/52, 0):

`h+p=-(1)/(52)`

As the directrix is x = 1/52:

`h-p=(1)/(52)`

Add the two equations together to eliminate p, and solve for h:

`\begin{array}{crcccr}\vphantom{\frac12}&(h&+&p&=&-(1)/(52)\\+&(h&-&p&=&(1)/(52)\vphantom{\frac12}\\\cline{2-6}\vphantom{\frac12}&2h&&&=&0\end{array}`

Therefore, h = 0.

Substitute the found value of h into one of the equations and solve for p:

`0+p=-(1)/(52)\implies p=-(1)/(52)`

Substitute the values of h, k and p into the standard formula:

`\begin{aligned}(y-0)^2&=4\left(-(1)/(52)\right)(x-0)\\\\y^2&=-(1)/(13)x\end{aligned}`

Therefore, the equation of the parabola in standard form with the given focus and directrix is:

`\large\boxed{\boxed{y^2&=-(1)/(13)x}}`