Problem 16
Question
Find the equation of the parabola through the point \((-2,4)\) if its vertex is at the origin and its axis is along the \(x\)-axis. Make a sketch.
Step-by-Step Solution
Verified Answer
The equation of the parabola is \(y^2 = -8x\).
1Step 1: Understand the Parabola Details
The parabola has its vertex at the origin, which means the vertex is at the point \(0, 0\). Since the axis is along the \(x\)-axis, the parabola opens either to the right or left. We need the equation of the parabola in the standard form of \(y^2 = 4px\) or \(y^2 = -4px\).
2Step 2: Use Point to Determine Equation
We need to use the given point \((-2, 4)\) to find the correct 'p' value. For the equation \(y^2 = 4px\), substitute \(y = 4\)and \(x = -2\), yielding the equation \(16 = 4p(-2)\)which simplifies to \(16 = -8p\).
Solve for \(p\) to get \(p = -2\). So, the equation is \(y^2 = -8x\).
3Step 3: Draw the Parabola
For the sketch, plot the vertex at \(0,0\). Since the parabola opens to the left due to the negative sign in \(y^2 = -8x\), it moves towards negative \(x\)-values. The point \((-2, 4)\) lies on this curve, confirming the shape and direction.
Key Concepts
Vertex FormConic SectionsCoordinate Geometry
Vertex Form
The vertex form is a helpful way to write the equation of a parabola. When we know the vertex's position, it becomes easier to understand the parabola's shape. The standard vertex form is given by:
- \( y = a(x - h)^2 + k \)
- \( y = ax^2 \)
- or \( x = ay^2 \)
Conic Sections
Parabolas are special because they are one of the four types of conic sections. Conic sections are shapes that are formed by slicing a double cone with a plane. The four sections include:
- Circle
- Ellipse
- Parabola
- Hyperbola
Coordinate Geometry
Coordinate geometry, or analytic geometry, is all about plotting mathematical equations into graphs. This branch of mathematics uses numbers to describe geometrical shapes. For parabolas, coordinate geometry allows us to find equations based on given points and vertex positions using coordinates.
In the given problem, the parabola goes through the point
((-2,4)) with the vertex at the origin (0,0). By applying coordinate geometry, you substitute these coordinates into the general parabola equation. This helps find the best-fit equation. Coordinate geometry simplifies visualizing algebraic expressions, thus making it easy to sketch graphs, determine points of intersection, or find distances between points and shapes.
Other exercises in this chapter
Problem 16
Sketch the three-leaved rose \(r=2 \sin 3 \theta\), and find the area of the region bounded by it.
View solution Problem 16
In Problems \(1-32\), sketch the graph of the given polar equation and verify its symmetry (see Examples 1-3). \(r=5-3 \cos \theta\) (limaçon)
View solution Problem 17
In each of Problems 1-20, a parametric representation of a curve is given. (a) Graph the curve. (b) Is the curve closed? Is it simple? (c) Obtain the Cartesian
View solution Problem 17
Find the equation of the given central conic. Ellipse with a focus at \((-3,0)\) and a vertex at \((6,0)\)
View solution