Problem 23 - Precalculus

Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{y^{2}}{1}-\frac{x^{2}}{4}=1$$

Step-by-Step Solution

Answer

The center of the hyperbola is (0,0). The vertices are (0,1) and (0,-1). The foci are (0, \(\pm \sqrt{5}\)). The equations of the asymptotes are \(y = \pm \frac{1}{2}x\). You can sketch the hyperbola by first sketching the asymptotes and then drawing the curve so that it approaches these asymptotes.

Step 1: Identify the Center

The center of the hyperbola is given by the point (h, k) in its standard form. Here, h and k are both zero, so the center is at the origin, (0, 0).

Step 2: Find the Vertices

The given equation uses a square of 'a=1' for y and 'b=2' for x. So, the vertices of the hyperbola are at a distance 'a' from the center along the y-axis. Thus, the vertices are at the points (0, \(\pm\)1).

Step 3: Calculate the Foci

The foci are at a distance 'c' from the center, where \(c = \sqrt{a^{2} + b^{2}}\). In this exercise, our a and b are 1 and 2 respectively. So, the foci are at (0, \(\pm \sqrt{5}\)).

Step 4: Discover the Asymptotes' Equations

The equations of the asymptotes for a hyperbola in this form are given by \( y = \pm \frac{a}{b}x\). Here, a:=1 and b:=2, so our asymptotes have the equations \(y = \pm \frac{1}{2}x\).

Step 5: Sketch the Hyperbola

Now to sketch the hyperbola and its asymptotes on the same graph. Draw the vertical lines y=\(\pm\)1 to mark the vertices, and the diagonal lines y=\(\pm \frac{1}{2}x\) to represent the asymptotes. These two sets of lines intersect at the origin (the center), and the hyperbola should be drawn within the funnel created by the asymptotes, touching the vertices.

Key Concepts

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. Depending on the angle of intersection, the shape can be a circle, an ellipse, a parabola, or a hyperbola. Specifically, a hyperbola results when the plane cuts through both naps of the cone and the angle of the intersection is steeper than the cone's side. The hyperbola in the given exercise \( \frac{y^{2}}{1}-\frac{x^{2}}{4}=1 \) is one such shape. Notice that the equation of a hyperbola is characterized by the subtraction of two squared terms, with the y-term being positive in this case, indicating a vertical hyperbola.

Asymptotes of a Hyperbola

The asymptotes of a hyperbola are imaginary lines that the curve approaches but never touches. These lines give us a framework to sketch the hyperbola accurately. The hyperbola's arms extend indefinitely, getting closer and closer to these asymptotes without ever intersecting them. For the equation \( \frac{y^{2}}{1}-\frac{x^{2}}{4}=1 \), the asymptotes are found using the formula \( y = \pm \frac{a}{b}x \), resulting in \( y = \pm \frac{1}{2}x \). When drawing these lines through the center, they form an 'X' that divides the plane and guides the sketching of the hyperbola.

Vertices of a Hyperbola

The vertices of a hyperbola are points where the hyperbola intersects its transverse axis. The transverse axis is the segment that joins the two parts of the hyperbola and passes through its center. In the given equation, the vertices are located at a distance 'a' from the center along the y-axis. Since 'a' is 1, the vertices for our equation \( \frac{y^{2}}{1}-\frac{x^{2}}{4}=1 \) are at the points (0, -1) and (0, 1). These points are crucial as they mark the sharpest turn of the graph and are essential anchors when sketching the hyperbola.

Foci of a Hyperbola

The foci (singular: focus) of a hyperbola are two fixed points located along the transverse axis that are used in the formal definition of the curve. Specifically, a hyperbola is the set of all points in the plane where the difference in distances to the two foci is a constant. To find the foci in the hyperbola given by \( \frac{y^{2}}{1}-\frac{x^{2}}{4}=1 \), we use the formula \( c = \sqrt{a^{2} + b^{2}} \) where 'a' and 'b' are values from the standard equation of a hyperbola. In this case, with 'a' being 1 and 'b' being 2, the foci are found at the coordinates (0, -\sqrt{5}) and (0, \sqrt{5}). These points are essential for the accurate construction and deeper understanding of the hyperbola's geometry.