Problem 42 - Algebra and Trigonometry

Question

use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ (y-2)^{2}-(x+3)^{2}=5 $$

Step-by-Step Solution

Answer

The center of the hyperbola is (-3,2), vertices are (-3, 2+√5) and (-3, 2-√5), foci are (-3, 2 + √10) and (-3, 2 - √10), and the asymptotes have equations y = -x + 5 and y = x + 1.

Step 1: Identify and Write Down the Center Coordinates

From the formula \((y-h)^2-(x-k)^2=1\), it can be identified that the center \((h, k)\) of the hyperbola is (-3,2).

Step 2: Identify a and b

Since the equation has the form \(\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1\), this implies that \(a^2=5\) and \(b^2=5\) as well. Therefore, \(a=b=\sqrt{5}\).

Step 3: Placing Vertices

The vertices will be placed at a distance 'a' from the center along the direction of the variable whose square is positive. Here, variable y has the positive square. So, vertices will be at \( (h, k+a) = (-3, 2+ \sqrt{5}) \) and \((-3, 2-\sqrt{5})\).

Step 4: Identifying Asymptotes

The equations of the asymptotes for a hyperbola standard equation are given as \( y = k \pm \frac{a}{b}(x - h) \). Substituting the values of h,k,a, and b, the equations become \(y = 2 \pm \frac{\sqrt{5}}{\sqrt{5}} (x+3)\) which simplifies to \( y= 2 \pm (x+3) \). Therefore, the equations of the asymptotes are \( y = -x+5 \) and \( y = x+1\).

Step 5: Find the Foci

The foci can be found using the formula \(c = \sqrt{|a^2 + b^2|}\). Substituting the values \( a = b = \sqrt{5} \), the foci c becomes \( c = \sqrt{2*5} = \sqrt{10} \). Since the hyperbola opens upwards and downwards, the foci will be located at \((h, k + c)\) and \((h, k - c)\), which are \((-3, 2 + \sqrt{10})\) and \((-3, 2 - \sqrt{10})\).

Key Concepts

Vertices

In a hyperbola, the vertices are crucial points that signify the extent of the opening along each direction. For the hyperbola described by the equation \[ (y-2)^2 - (x+3)^2 = 5 \], the vertices provide insight into its orientation. They lie along the axis associated with the positive term in the equation.

Since the positive square in our equation involves the variable \(y\), the hyperbola opens vertically (up and down). The vertices are computed by moving a distance of \( \sqrt{5} \) from the center along the y-axis. In this case, with a calculated center at \((-3, 2)\) and \(a = \sqrt{5}\), the vertices are found at:

• \((-3, 2 + \sqrt{5})\)
• \((-3, 2 - \sqrt{5})\)

These points indicate the "top" and "bottom" tips of the hyperbola, as they stretch away from the center.

Asymptotes

Asymptotes are lines that the hyperbola approaches infinitely but never actually touches. They act like guiding borders, sketching the general direction and shape of the hyperbola.

In mathematical terms, the asymptotes for a hyperbola with a vertical opening like ours have the form:\[ y = k \pm \frac{a}{b}(x - h) \]By substituting the determined values \(h = -3\), \(k = 2\), and both \(a\) and \(b = \sqrt{5}\), the asymptotes simplify into:

• \( y = -x + 5 \)
• \( y = x + 1 \)

These equations help in accurately sketching the hyperbola, offering a clear direction on how sharply it opens and closes.

Foci

The foci of a hyperbola offer significant insight into its structure, determining its aperture. The foci are located further outside the vertices, and are essential for understanding the hyperbola's elongation.

They are typically calculated using the formula:\[ c = \sqrt{|a^2 + b^2|} \] For our hyperbola, insert \(a = \sqrt{5}\) and \(b = \sqrt{5}\) to calculate:\( c = \sqrt{10} \).
For an up and down opening hyperbola like ours, the foci locations are:

• \((-3, 2 + \sqrt{10})\)
• \((-3, 2 - \sqrt{10})\)

These points express where the hyperbola curves converge. Thus, the hyperbola looks more like a wide ellipse between these focal points, stretched vertically along the y-axis.

Equation of Hyperbola

The equation of a hyperbola is what fundamentally describes its structure and geometry. It takes a specific form for varied orientations, and understanding this is key to graphing it accurately.

For the hyperbola with the equation \((y-2)^2 - (x+3)^2 = 5\), it implies a vertically oriented hyperbola since \((y-k)^2\) is the positive term. We can rewrite the equation in its standard form:\[ \frac{(y-2)^2}{5} - \frac{(x+3)^2}{5} = 1 \] Here, we identify \(a^2 = 5\) and \(b^2 = 5\), reflecting equal semi-axis lengths.
This equation provides all the necessary parameters to determine features like the vertices, foci, and asymptotes, and hence, effectively "governs" the hyperbola's graph.