Problem 6
Question
Find the coordinates of the focus and the equation of the directrix for each parabola. Make a sketch showing the parabola, its focus, and its directrix. $$y^{2}+3 x=0$$
Step-by-Step Solution
Verified Answer
Focus: \((-\frac{3}{4}, 0)\), Directrix: \(x = \frac{3}{4}\).
1Step 1: Identify the type of parabola
The given equation of the parabola is \(y^2 + 3x = 0\). Notice that it is of the form \(y^2 = -4px\), which is a standard equation for a parabola that opens to the left or right.
2Step 2: Rearrange the equation
Rewrite the equation \(y^2 + 3x = 0\) in the form \(y^2 = -4px\). We get \(y^2 = -3x\). Here, we compare this with \(y^2 = -4px\) to find \(-4p = -3\), giving \(p = \frac{3}{4}\).
3Step 3: Determine the orientation of the parabola
Since the equation \(y^2 = -3x\) fits the form \(y^2 = -4px\) and \(-4p\) is negative, the parabola opens to the left.
4Step 4: Find the focus of the parabola
For a parabola \(y^2 = -4px\), the focus is located at \((-p, 0)\) if it opens to the left. Substituting the value of \(p\), the focus is at \(\left(-\frac{3}{4}, 0\right)\).
5Step 5: Find the directrix of the parabola
The equation of the directrix for a parabola \(y^2 = -4px\) is \(x = p\). Thus, the directrix is \(x = \frac{3}{4}\).
6Step 6: Sketch the parabola, focus, and directrix
Draw the parabola on the Cartesian plane with its vertex at the origin \((0, 0)\), opening to the left. Place the focus at \((-\frac{3}{4}, 0)\) and draw the line for the directrix at \(x = \frac{3}{4}\). The vertex, focus, and directrix all align on the horizontal axis.
Key Concepts
Focus and DirectrixParabola OrientationParabola SketchingParabola Equations
Focus and Directrix
In parabolas, the concepts of "focus" and "directrix" are pivotal to understanding their properties. The focus is a fixed point inside the parabola, and every point on the parabola is equidistant from both the focus and the directrix, which is a fixed line. This relationship is crucial in describing how parabolas curve.
- The focus acts like a guiding point that the parabola wraps around.
- The directrix serves as a boundary on the opposite side of the curve from the focus.
Parabola Orientation
Understanding parabola orientation is necessary to graph its curriculum accurately. The orientation describes the direction in which the parabola opens. This is determined by the sign and placement of terms in its equation.
- If the term \(y^2\) appears, the parabola opens horizontally either to the left or right.
- If the term \(x^2\) appears, it opens vertically, either upwards or downwards.
- The direction can further be determined by the sign:
- A positive term like \(y^2 = 4px\) opens to the right.
- A negative term like \(y^2 = -4px\) opens to the left.
Parabola Sketching
Sketching a parabola involves illustrating its key elements correctly on a coordinate plane. Start by identifying the vertex, focus, and directrix.
- Vertex: This is typically located at the origin \((0, 0)\) for standard equations unless shifted by additional transformations.
- Focus: Place it as calculated, here at \((-\frac{3}{4}, 0)\), guiding the curve of the parabola.
- Directrix: Draw it at \(x = \frac{3}{4}\), ensuring it is perpendicular to the axis through the vertex.
Parabola Equations
Equations of parabolas are mathematical representations that describe their curve. The standard form of a parabola equation varies depending upon its orientation.
- Vertical Parabolas: Typically use the form \((x - h)^2 = 4p(y - k)\), where \((h, k)\) is the vertex.
- Horizontal Parabolas: Use the form \((y - k)^2 = 4p(x - h)\).
Other exercises in this chapter
Problem 6
\(r=3+3 \sin \theta\)
View solution Problem 6
In Problems \(1-32\), sketch the graph of the given polar equation and verify its symmetry (see Examples 1-3). \(r=4 \sin \theta\)
View solution Problem 7
Name the conic corresponding to the given equation. \(9 x^{2}+4 y^{2}=9\)
View solution Problem 7
Name the conic or limiting form represented by the given equation. Usually you will need to use the process of completing the square. $$ y^{2}-5 x-4 y-6=0 $$
View solution