Problem 18 - College Algebra with Corequisite Support

Question

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. \((y-2)^{2}=\frac{4}{5}(x+4)\)

Step-by-Step Solution

Answer

Vertex \((-4, 2)\), Focus \((-\frac{19}{5}, 2)\), Directrix \(x = -\frac{21}{5}\).

Step 1: Identify the given form

The given equation is \[(y-2)^2 = \frac{4}{5}(x+4).\] This equation is a parabola in the form \((y-k)^2 = 4p(x-h)\), which means it opens sideways.

Step 2: Rewrite the equation in standard form

The equation already resembles the standard form of a parabola \[(y-k)^2 = 4p(x-h),\] where \(h = -4\) and \(k = 2.\) Thus, there's no need to rewrite it further as it clearly shows the form \((y-k)^2 = 4p(x-h).\)

Step 3: Determine the vertex

The vertex \(V\) of the parabola \[(y-2)^2 = \frac{4}{5}(x+4)\] is given by the point \((h, k)\). Here, \(h = -4\) and \(k = 2\), so the vertex is \(V = (-4, 2).\)

Step 4: Find the parameter p

The term \(4p\) in \[(y-k)^2 = 4p(x-h)\] corresponds to the coefficient of \((x+4)\), which is \(\frac{4}{5}.\) Solving for \(p\), we have \[4p = \frac{4}{5} \Rightarrow p = \frac{1}{5}.\]

Step 5: Determine the focus

The focus \(F\) is determined using \(p.\) For a parabola opening sideways in the form \((y-k)^2 = 4p(x-h)\), the focus is at \((h + p, k).\) So, with \(h = -4\), \(p = \frac{1}{5}\), and \(k = 2\), the focus is \[F = \left(-4 + \frac{1}{5}, 2\right) = \left(-\frac{19}{5}, 2\right).\]

Step 6: Determine the directrix

The directrix of the parabola is a vertical line given by \[x = h - p.\] Substituting \(h = -4\) and \(p = \frac{1}{5},\) we find\[x = -4 - \frac{1}{5} = -\frac{21}{5}.\] Therefore, the equation of the directrix is \(x = -\frac{21}{5}.\)

Key Concepts

Vertex of a Parabola

The vertex is a key point of a parabola, acting as its turning point or the point where the curve is the closest or farthest to its directrix or axis of symmetry. For the equation \[(y-k)^2 = 4p(x-h),\]the vertex \((h, k)\) is easily identified.

• It's the point where the parabola changes direction, moving either left or right when the parabola opens sideways.
• In the equation \[(y-2)^2 = \frac{4}{5}(x+4),\]the vertex is given by \(h = -4\) and \(k = 2\), which means the vertex is \(V = (-4, 2)\).

Having the vertex helps in graphing the parabola accurately on a coordinate plane.
It also acts as a reference point for finding other features like the focus and directrix.

Focus of a Parabola

The focus of a parabola is a critical point, lying inside the parabola, that helps define its shape and orientation. Every point on the parabola is equidistant from the focus and the directrix. For a parabola given by \[(y-k)^2 = 4p(x-h),\]the focus is at \((h + p, k)\).

• In our given example, \(p\) is \(\frac{1}{5}\), making the coordinate of the focus \(F = \left(-\frac{19}{5}, 2\right)\).
• The focus is always placed along the axis of symmetry of the parabola, thus influencing its direction.

Understanding the focus helps in determining how narrow or wide a parabola appears, and it's a point that reflects the properties of the parabola in relation to the directrix.

Directrix of a Parabola

The directrix is an integral component of the definition of a parabola. It is a fixed line that, together with the focus, helps determine the set of points forming the parabola. For a parabola represented by \[(y-k)^2 = 4p(x-h),\]the directrix is a vertical line given by \(x = h - p\).

• It is positioned opposite to the direction in which the parabola opens. This means that while the focus sits inside the parabola, the directrix will be outside.
• From our equation, \(h = -4\) and \(p = \frac{1}{5}\), so the directrix is at \(x = -\frac{21}{5}\).

The directrix serves as an imaginary boundary line that complements the position of the focus, ensuring the symmetry and shape of the parabola. Each point on the parabola is equidistant from the focus and directrix, defining the curve's path in the coordinate plane.