Problem 26 - 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. \(y^{2}-24 x+4 y-68=0\)

Step-by-Step Solution

Answer

Vertex: \((-3, -2)\), Focus: \((3, -2)\), Directrix: \(x = -9\)

Step 1: Rewrite the Equation

First, rewrite the given equation by moving the terms involving \(x\) to one side and the rest to the other side. The original equation is: \[ y^2 - 24x + 4y - 68 = 0 \]Rearrange it as follows:\[ y^2 + 4y = 24x + 68 \]

Step 2: Complete the Square

Complete the square for the \(y\)-terms on the left. Take the coefficient of \(y\), which is 4, divide it by 2, and square the result. \[ \text{Coefficient of } y = 4, \text{ divide by 2 and square: } \left(\frac{4}{2}\right)^2 = 4 \]Add and subtract 4 on the left to complete the square:\[ y^2 + 4y + 4 = 24x + 68 + 4 \]This simplifies to:\[ (y+2)^2 = 24x + 72 \]

Step 3: Convert to Standard Form

The equation \((y+2)^2 = 24x + 72\) must be rewritten in parabola's standard form \((y-k)^2 = 4p(x-h)\).Subtract 72 from both sides: \[ (y+2)^2 = 24(x + 3) \]Now the equation is in the standard form \((y-k)^2 = 4p(x-h)\) with \(h = -3\), \(k = -2\), and \(4p = 24\). So, \(p = 6\).

Step 4: Determine the Vertex

The vertex \(V\) of the parabola in the standard form \((y-k)^2 = 4p(x-h)\) is at \((h, k)\). Since \(h = -3\) and \(k = -2\), the vertex is: \[ V = (-3, -2) \]

Step 5: Find the Focus

The focus \(F\) of the parabola is \(p\) units away from the vertex along the axis of symmetry. For \((y-k)^2 = 4p(x-h)\), the axis is horizontal and the focus is \((h+p, k)\).Thus, \(p = 6\), so the focus is:\[ F = (-3+6, -2) = (3, -2) \]

Step 6: Determine the Directrix

The directrix \(d\) of the parabola \((y-k)^2 = 4p(x-h)\) is a vertical line given by \(x = h - p\).Here, \(h = -3\) and \(p = 6\), so the directrix is:\[ d: x = -3 - 6 = -9 \]

Key Concepts

Vertex

In the world of parabolas, the vertex plays a critical role. It serves as the turning point of the parabola where the curve changes direction. In the standard form of a parabola centered on the x-axis, the equation looks like y-k)^2 = 4p(x-h . Here, (h, k is the vertex. This point is crucial as it gives us a reference for determining other properties, such as the focus and directrix.

• The vertex coordinates are found directly from the standard equation in the format y-k)^2 = 4p(x-h .
• For the given equation y+2)^2 = 24(x + 3 , the vertex is (-3, -2 .
• Remembering this formula helps simplify identifying the vertex as it is simply translating the constant terms inside the parentheses.

With the vertex identified, it helps in sketching the parabola on a graph. By understanding the vertex, you can better determine how the parabola sits on the coordinate plane.

Focus

The focus of a parabola is a special point that, along with the directrix, defines the shape of the parabola. It holds the property that every point on the parabola is equidistant from the focus and the directrix. This unique feature makes the focus essential in defining the characteristic sharpness of the curve. For a parabola in the standard form y-k)^2 = 4p(x-h , the focus sits at (h+p, k , which results from moving