Problem 44 - Contemporary Precalculus

Question

In Exercises \(43-54\), find the equation of the parabola satisfying the given conditions. Vertex (0,1)\(;\) axis \(x=0 ;(2,-7)\) on graph.

Step-by-Step Solution

Answer

Answer: The equation of the parabola is $x^2=-\frac{1}{2}(y-1)$.

Step 1: Plug the Given Point into the Equation

Plug the coordinates of the given point (2,-7) into the equation: $$(2)^2=4p(-7-1)$$

Step 2: Solve for p

Solve the equation for p: $$4=4p(-8)$$ $$4=-32p$$ $$p=-\frac{1}{8}$$

Step 3: Write the Equation of the Parabola

Replace p in the equation with the value we found: $$x^2=4\left(-\frac{1}{8}\right)(y-1)$$ $$x^2=-\frac{1}{2}(y-1)$$

Step 4: Final Equation of the Parabola

The equation of the parabola satisfying the given conditions is: $$x^2=-\frac{1}{2}(y-1)$$

Key Concepts

Vertex Form of a Parabola

Understanding the vertex form of a parabola is crucial when dealing with quadratic functions. The vertex form is written as \(y - k = a(x - h)^2\) where \(h, k\) is the vertex of the parabola, and the value of \(a\) determines the width and direction of the opening.

If \(a > 0\), the parabola opens upwards, and if \(a < 0\), it opens downwards. The absolute value of \(a\) affects the 'stretch' of the parabola — the smaller the value, the wider the parabola; the larger the value, the more 'narrow' or 'steep' it appears. For example, in the final equation from our exercise, \(x^2 = -\frac{1}{2}(y - 1)\), flipping the equation into the vertex form gives us \(y - 1 = -2x^2\), which shows a parabola with vertex at \(0, 1\) that opens downward due to the negative coefficient.

Directrix and Focus

The directrix and focus are defining features of a parabola. The focus is a fixed point on the interior of the parabola, and the directrix is a line perpendicular to the axis of symmetry that the parabola does not intersect.

Every point on the parabola is equidistant to the focus and the directrix. This constant distance is denoted by the variable \(p\). In our original exercise, once we solve for \(p\), that value helps define the directrix (\(y = k + p\) if the parabola opens upwards or downwards, and \(x = h + p\) if it opens sideways) and the coordinates of the focus (\(h, k + p\) or \(h + p, k\))

Graphing Parabolas

Graphing a parabola requires understanding its vertex, axis of symmetry, and the direction it opens. The vertex form of the equation simplifies finding the vertex and the axis of symmetry, which is a vertical line through the vertex.

For our given problem, the vertex is at \(0, 1\), and since \(x=0\), that's also the axis of symmetry. When graphing, we mark the vertex, plot additional points, and sketch the curve, ensuring it's equidistant on both sides of the axis. Points like \(2, -7\) help us understand the direction and width of the parabola. Remember, the parabola is symmetric, so plotting points on one side of the vertex helps you mirror them on the other side.

Solving for p in Parabolas

In the context of parabolas, \(p\) represents the distance from the vertex to the focus and equals the distance from the vertex to the directrix. Solving for \(p\) is essential to complete the description of the parabola's geometry.

In our example solution, plugging the point \(2, -7\) into the parabola's equation \(x^2=4p(y-k)\) and solving for \(p\) gave us \(p=-\frac{1}{8}\). This indicates the focus is \(\frac{1}{8}\) units below the vertex since \(p\) is negative, and the parabola opens downwards. Understanding \(p\) allows us to graph the parabola accurately and define its directrix and focus for a complete graphical representation.