Problem 41 - Precalculus with Calculus Previews

Question

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Center \((-1,3),\) one vertex \((-1,4),\) passing through \((-5,3+\sqrt{5})\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \((y - 3)^2 - \frac{(x + 1)^2}{4} = 1\).

Step 1: Understanding the Hyperbola Structure

A hyperbola centered at \((h, k)\) with a vertical transverse axis follows the equation: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]Here, the center is given as \((-1, 3)\), so \(h = -1\) and \(k = 3\).

Step 2: Determining 'a'

The vertex of the hyperbola is given as \((-1, 4)\). Since the center is \((-1, 3)\), the distance from the center to a vertex vertically is \(1\). Therefore, \(a = 1\) and \(a^2 = 1\).

Step 3: Substitute Known Values

Substitute \(h = -1\), \(k = 3\), and \(a^2 = 1\) into the equation for the hyperbola:\[ \frac{(y - 3)^2}{1} - \frac{(x + 1)^2}{b^2} = 1 \]Simplify this to:\[ (y - 3)^2 - \frac{(x + 1)^2}{b^2} = 1 \]

Step 4: Use Given Point to Find 'b'

The hyperbola passes through \((-5, 3+\sqrt{5})\). Substitute this point into the equation:\[ ((3+\sqrt{5}) - 3)^2 - \frac{((-5) + 1)^2}{b^2} = 1 \]This becomes:\[ (\sqrt{5})^2 - \frac{4^2}{b^2} = 1 \]\[ 5 - \frac{16}{b^2} = 1 \]

Step 5: Solve for 'b^2'

Subtract \(1\) from \(5\) to isolate the term involving \(b^2\):\[ 4 = \frac{16}{b^2} \]Multiply both sides by \(b^2\):\[ 4b^2 = 16 \]Divide by \(4\) to solve for \(b^2\):\[ b^2 = 4 \]

Step 6: Write the Equation of the Hyperbola

Using the values \(h = -1\), \(k = 3\), \(a^2 = 1\), and \(b^2 = 4\), the equation of the hyperbola is:\[ \frac{(y - 3)^2}{1} - \frac{(x + 1)^2}{4} = 1 \] Simplify this to:\[ (y - 3)^2 - \frac{(x + 1)^2}{4} = 1 \]

Key Concepts

Conic Sections

Conic sections are curves that arise from slicing a double cone with a plane. Imagine holding two ice cream cones bottom to bottom; the resulting shape can be sliced in different ways to produce these curves. The main types of conic sections are:

• Circles - slices parallel to the base of the cone, forming a perfectly round cross-section.
• Ellipses - slices at an angle, but not too steeply. They look like stretched circles.
• Parabolas - slices parallel to the side of the cone, forming a U-shaped curve.
• Hyperbolas - slices that are more steep, cutting through both halves of the double cone and forming two oppositely curved sections.

Hyperbolas, the focus of our original exercise, appear as two separate curves known as branches, which open in opposite directions. These are distinct from other conic sections by their equation, which involves subtraction, shown by their basic form: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \] This structure signifies that the hyperbola's general shape will open vertically.

Equation of Hyperbola

The equation of a hyperbola is fundamental in understanding its geometric properties. Hyperbolas have two standard forms, depending on the orientation of the transverse axis, which is the axis along which the branches of the hyperbola open:

• Vertical Transverse Axis: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]In this, the hyperbola opens up and down, centered at \((h, k)\).
• Horizontal Transverse Axis: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]Here, the branches open left and right, with the same center point.

In our exercise, we used the equation for a vertical transverse axis because the center was given, along with the vertical vertex.
Knowing the center \((-1, 3)\) and one vertex \((-1, 4)\), we knew there was only a vertical distance, indicating the transverse axis is vertical and \(a = 1\).
Depending on whether \(a\) or \(b\) are larger, a hyperbola can appear more squished or elongated along its different axes.

Vertices of Hyperbola

The vertices of a hyperbola are significant points on the curve where the branches are closest or furthest from the center. For a hyperbola with a vertical transverse axis, the vertices are directly above and below the center point. The formula for the vertices when the center is \((h, k)\) is very straightforward:

• Vertical transverse axis: Vertices at \((h, k \pm a)\)
• Horizontal transverse axis: Vertices at \((h \pm a, k)\)

In our problem, the center was \((-1, 3)\), and a vertex was given as \((-1, 4)\), which indicates that \(a = 1\) (the distance from the center to the vertex along the transverse axis). Therefore, the other vertex would be at \((-1, 2)\).This understanding of vertices helps in sketching the shape of the hyperbola and defining the size of its central rectangle, an auxiliary rectangle that aids in the graphing and visualization of the hyperbola's branches.