Problem 19 - Algebra and Trigonometry

Question

\(11-22\) . Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$ x^{2}+6 y=0 $$

Step-by-Step Solution

Answer

Focus: \(\left(0, -\frac{3}{2}\right)\); Directrix: \(y = \frac{3}{2}\); Focal diameter: 6.

Step 1: Identify the general form of the parabola

The given equation is \(x^2 + 6y = 0\). For a parabola that opens vertically, the equation is generally in the form \((x-h)^2 = 4p(y-k)\). Here, the equation can be rewritten as \(x^2 = -6y\).

Step 2: Compare with the standard form

Compare \(x^2 = -6y\) with the standard form \((x-h)^2 = 4p(y-k)\). From this, we find that \((h, k) = (0, 0)\) and \(4p = -6\).

Step 3: Calculate the value of p

We solve for \(p\) in the equation \(4p = -6\). This gives \(p = -\frac{6}{4} = -\frac{3}{2}\).

Step 4: Determine the focus

For a parabola of the form \((x-h)^2 = 4p(y-k)\), the focus is at \((h, k+p)\). Since \((h, k) = (0, 0)\) and \(p = -\frac{3}{2}\), the focus is \(\left(0, -\frac{3}{2}\right)\).

Step 5: Determine the directrix

The directrix of the parabola \((x-h)^2 = 4p(y-k)\) is given by the line \(y = k-p\). So, the directrix is \(y = 0 + \frac{3}{2} = \frac{3}{2}\).

Step 6: Determine the focal diameter

The focal diameter (length of the latus rectum) is given as the absolute value of \(4p\). Here, it is \(|-6| = 6\).

Step 7: Sketch the graph

The parabola \(x^2 = -6y\) opens downwards because the coefficient of \(y\) is negative. The vertex is at the origin \((0,0)\), the focus is at \(\left(0, -\frac{3}{2}\right)\), and the directrix is the line \(y = \frac{3}{2}\). Draw the parabola opening downward, ensuring it is symmetrical about the y-axis.

Key Concepts

Focus of a Parabola

The focus of a parabola is a point located inside the curve where all the lines reflecting off the inner surface converge. It plays a crucial role in the geometric properties of a parabola. For a vertical parabola, expressed in the standard form \((x-h)^2 = 4p(y-k)\), the focus can be pinpointed at the coordinate \((h, k+p)\).
In this specific exercise, the equation is rewritten as \(x^2 = -6y\) with

• \((h, k) = (0, 0)\)
• \(p = -\frac{3}{2}\)

Thus, the focus of the parabola is found at \(\left(0, -\frac{3}{2}\right)\). The focus's position significantly affects the parabolic shape and direction. Here, since \(p\) is negative, the parabola opens downward.

Directrix of a Parabola

The directrix of a parabola is an essential line that helps guide the shape of the parabola, lying opposite from the vertex as compared to the focus. It maintains a constant distance from any point on the parabola. For a parabola with a vertical axis, the equation of the directrix is given by the formula:

• \(y = k-p\)

In this exercise, substituting the values obtained:

• \(k = 0\)
• \(p = -\frac{3}{2}\)

We derive the equation of the directrix as \(y = \frac{3}{2}\). The alignment between the directrix and the focus essentially dictates the symmetry of the parabola.

Focal Diameter

Focal diameter, or the latus rectum, is a measure of the width of the parabola at the level of the focus. This is not just any width, but the width which passes through the focus and is perpendicular to the axis of symmetry. It's an important parameter because it reflects how 'wide' or 'narrow' the parabola is around its focus.The focal diameter is

• given as the absolute value of \(4p\)

In this question, we compute it to be:

• \(|4p| = |-6| = 6\).

This value indicates the total length of the line segment passing through the focus and parallel to the directrix. Understanding the focal diameter helps to appreciate not just the points of convergence but the spread of the parabola in a given direction.

Vertex of a Parabola

The vertex of a parabola is the key point that either stands at the foremost tip or bottom depending on the parabola's direction. It acts as the fulcrum around which the parabola opens either upwards or downwards.In our particular equation, \(x^2 + 6y = 0\), when it is rewritten in the form \((x-h)^2 = 4p(y-k)\), we identify

• \((h, k) = (0, 0)\)

This directly gives us the vertex point at \((0,0)\). From this point, the curve of the parabola sweeps downward as confirmed by the negative coefficient of \(y\).
The vertex being on the origin also implies a symmetrical structure about the y-axis, biologically defining the shape's inception.