Understanding the Directrix and Focus of a Parabola y8x^2-1
When dealing with the equation of a parabola, it's essential to understand the vertex form of the equation for simplicity. In the case of our parabola y8x^2-1, we can derive the vertex, directrix, and focus with a series of carefully structured steps. In this article, we'll explore how to find the directrix and focus of the parabola.
Equation and Vertex
The given equation is y8x^2-1, which is already in vertex form ya(x-h)^2 k, where the vertex (h, k) is (0, -1). This tells us that the vertex of our parabola is at ((-2, -1)).
Shifting the Parabola for Simplification
For simplicity, we can temporarily shift the parabola so its vertex is at the origin (0, 0). This is done by substituting (x' x 2) and (y' y 1) into the equation.
The equation then becomes:
[ (y' 1) 8(x' - 0)^2 - 1 implies y' 8x'^2]Determining the Directrix and Focus
The equation (y' 8x'^2) is now much simpler. We know from the general form of a parabola (x^2 4py), the focus is at ((0, p)) and the directrix is (y -p).
From the equation (y' 8x'^2), we can derive:
[ 4p 8 implies p 2]This means the focus of the temporary parabola is at ((0, 2)) and the directrix is (y -2).
Translating the Information Back
Now, we need to translate this information back to the original coordinates, shifting the parabola two units to the left and one unit down. This means the focus and directrix should be adjusted accordingly.
Focus: ((0, 2) - (2, -1) (-2, 1/32 - 1) (-2, -31/32))
Directrix: (y -2 -1 -33/32)
Conclusion
By following these steps, we have found that the directrix of the original parabola (y 8x^2 - 1) is (y -33/32), and the focus is at ((-2, -31/32)). This process of temporarily shifting the parabola, determining the focus and directrix, and then translating back is a powerful method for understanding and manipulating parabolas in the equation domain.
Frequently Asked Questions
Q: How can I find the vertex of a parabola?
A: The vertex of a parabola in the form (y ax^2 bx c) is given by the formula (left(-frac{b}{2a}, text{substitute } x -frac{b}{2a} text{ to find } yright)). For (y 8x^2 - 1), the vertex is at ((0, -1)).
Q: What is the difference between the directrix and focus of a parabola?
A: The directrix is a line perpendicular to the axis of symmetry and equidistant from the focus. For a parabola (x^2 4py), the focus is at ((0, p)) and the directrix is (y -p).