Introduction

Determining the focus and vertex of a parabola is a common task in mathematics, particularly in calculus and geometry. This guide provides a step-by-step solution to find the focus and vertex of the parabola described by the equation xy^2 - 8x - 8y 0.

Revisiting the Given Equation

The given equation of the parabola is:

xy^2 - 8x - 8y 0

We first aim to simplify and rearrange this equation into a more recognizable form. Let's start by isolating the terms involving x and y:

Rearranging the Equation

Starting with the given equation:

xy^2 - 8x - 8y 0

We can factor out common terms to get:

xy^2 8x - 8y

To further simplify, we can rewrite this as:

xy^2 8(x - y)

Let's now introduce a new variable z x - y. Substituting this into the equation, we get:

yz^2 8z

Rearrange it to:

z^2 8z - y

Further simplifying, we have:

z^2 - 8z y 0

Now, let's express y in terms of z:

y z^2 - 8z

Identifying the Vertex

To find the vertex, we complete the square for the equation in terms of z:

y z^2 - 8z

Completing the square, we get:

y (z - 4)^2 - 16

This shows that the vertex is at z 4, y -1. Substituting back z x - y into these values:

x - y 4 thus y 4 - x

Therefore, the vertex in terms of x, y coordinates is:

(4, 0)

Identifying the Focus

The standard form of the parabola opens upward/downward and we can identify the focus.

The general form of a parabola that opens upwards is:

(x - h)^2 4p(y - k)

Comparing it with our equation, we have:

h 4, k 0, p 1/4

Since the coefficient of (z - 4)^2 is 1/16, p 1/16. The focus of a parabola that opens upwards is at a distance p from the vertex. Since our vertex is at (4, 0) and p 1:

The focus is at:

(4, 0 1) (4, 1)

Summary

- Vertex: (4, 0)

- Focus: (4, 1)

Alternative Method: Analyzing Directrix and Focus

A parabola is also defined by its focus, directrix, and vertex. Given the equation xy^2 - 8x - 8y 0, we can identify the directrix and focus by logical analysis.

The focus, as previously determined, is (4, 1).

The directrix, perpendicular to the axis of symmetry and equidistant from the focus, is given by:

y x - 2

The point on the curve that is equidistant from the focus and the directrix is:

P(8, 0)

Thus, the equation for the directrix to access the focus and the point on the curve is as follows:

Distance of P from F Distance of P from directrix

(8 - 1)2 (0 1)2 (8 - m)2 (-1 - m)2

65 (8 - m)2 (1 m)2

Solving this will give the values for the directrix and the point 8, 0 on the curve.

Conclusion

By solving the given equation and completing the square, we successfully identified the vertex and focus of the parabola described by xy^2 - 8x - 8y 0. This detailed guide can be applied to similar problems to find the focus and vertex of a parabola.