Problem 42 - College Algebra

Question

Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(y-1)^{2}=-8 x$$

Step-by-Step Solution

Answer

The vertex of the parabola is at (0, 1), the focus is at (-2, 1), and the equation of the directrix is x = 2.

Step 1: Rewrite equation in standard form

The general form of a parabola equation is \( (y-k)^{2} = 4a(x-h) \) where (h, k) is the vertex. Therefore, rewrite the given equation \( (y-1)^{2}=-8x \) into standard form.

Step 2: Identify the vertex

Identify the vertex (h, k) which is the point (0, 1) in the equation.

Step 3: Find the value of a

Relate the coefficient of x to 4a. In equation \( (y-1)^{2}=-8x \), 4a is -8 so a = -2.

Step 4: Identify the focus

The focus of a parabola \( (y-k)^{2} = 4a(x-h) \) is \( (h+a, k) \). So, with a = -2 and the vertex (0, 1), the focus is (-2, 1).

Step 5: Identify the directrix

The equation of the directrix of a parabola \( (y-k)^{2} = 4a(x-h) \) is \( x = h-a \). So the equation of the directrix is x = 0 - (-2) or x = 2.

Step 6: Graph the parabola

The parabola opens towards the left because a is negative. The vertex is the highest point on the graph at (0, 1). Plot this point, along with the focus at (-2, 1), and draw the directrix line at x = 2. Then create the shape of the parabola.