Problem 56 - Precalculus: Mathematics for Calculus

Question

Complete the square to determine whether the graph of the equation is an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an cllipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$x^{2}-y^{2}=10(x-y)+1$$

Step-by-Step Solution

Answer

The equation represents a hyperbola centered at (5,5) with vertices (4,5) and (6,5).

Step 1: Rearrange the Equation

Start by rewriting the given equation \(x^{2}-y^{2}=10(x-y)+1\) such that all terms are on one side and it equals zero. Distribute and group similar terms: \(x^2 - y^2 - 10(x - y) = -1\). This simplifies to \(x^2 - y^2 - 10x + 10y + 1 = 0\).

Step 2: Complete the Square for x-terms

To complete the square with respect to \(x\), take the \(x\) terms \(x^2 - 10x\). The coefficient of \(x\) is \(-10\), half of \(-10\) is \(-5\), and squaring \(-5\) gives \(25\). Add and subtract \(25\) from the equation to obtain \((x - 5)^2 - 25\).

Step 3: Complete the Square for y-terms

Now complete the square for the \(y\) terms \(-y^2 + 10y\). Factor out \(-1\) giving \(-(y^2 - 10y)\). The coefficient of \(y\) is \(10\), half of \(10\) is \(5\), and \(5^2\) is \(25\). Add and subtract \(25\): it becomes \(-(y - 5)^2 + 25\).

Step 4: Simplify and Analyze the Equation Type

Substitute the completed squares back into the equation: \((x-5)^2 - 25 - ( (y-5)^2 - 25 ) + 1 = 0\). Simplify to get \((x-5)^2 - (y-5)^2 = 1\). This is the standard form of a hyperbola, \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), indicating the graph represents a hyperbola.

Step 5: Find the Hyperbola Parameters

From the equation \((x-5)^2 - (y-5)^2 = 1\), conclude that the center is at \((5, 5)\), \(a^2 = 1\), \(b^2 = 1\). The vertices are at \((5\pm 1, 5)\), i.e., \((4, 5)\) and \((6, 5)\). The length of the transverse axis is \(2a = 2\). To find the foci, compute \(c^2 = a^2 + b^2 = 2\), so \(c = \sqrt{2}\). The foci are \((5\pm \sqrt{2}, 5)\).

Step 6: Asymptotes of the Hyperbola

The asymptotes of a hyperbola \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\) are given by \(y-k = \pm \frac{b}{a}(x-h)\). Here, \(a = b = 1\) so the asymptotes are \(y - 5 = \pm (x - 5)\) or \(y = x\) and \(y = -x + 10\).

Step 7: Sketch the Hyperbola

Draw the hyperbola with center at \((5,5)\), marking the vertices, \((4,5)\) and \((6,5)\), and foci \((5\pm \sqrt{2},5)\). Sketch the transverse axis (parallel to the x-axis), and draw the asymptotes \(y = x\) and \(y = -x + 10\) as guidelines for the hyperbola’s arms.

Key Concepts

Ellipse

An ellipse is a distinct type of conic section that appears as an elongated circle. Think of it like stretching a circle along a particular direction, either horizontally or vertically.

Key features of an ellipse include:

• Two focal points (foci) that determine its shape.
• A major axis (the longest diameter) and a minor axis (the shortest diameter).
• The center, which is the midpoint between the foci.

The equation of an ellipse in its standard form is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where

• \(h, k\) sets the center coordinates.
• \(a^2\) and \(b^2\) determine the lengths of the axes.

If you were to graph the equation and find that both the x-term and y-term are positive, and the sum is 1 when rewritten in the standard form, then you've pinpointed an ellipse.

In solving a conic section problem, discovering these attributes helps in sketching and understanding an ellipse's presence and characteristics.

Parabola

A parabola is a unique conic section that looks like a U or an inverted U, depending on its orientation. Notably, it only stretches in one direction.

Key characteristics of a parabola include:

• A singular focus, which is a point inside the curve.
• A directrix, which is a line outside the curve.
• A vertex, which is the point where the parabola turns, letting the curve open widest from that point.

The standard form equation of a parabola is:- For vertical parabolas: \[ (x-h)^2 = 4p(y-k) \]- For horizontal parabolas:\[ (y-k)^2 = 4p(x-h) \]where
- \(h, k\) mark the vertex coordinates, and - \(p\) indicates the distance from the vertex to the focus (or the directrix).Parabolas are particularly crucial in physics and engineering due to their focusing properties. When provided with an equation, analyzing the squared term tells you if a graph represents a parabola, implying its direction and orientation.

Hyperbola

A hyperbola is another type of conic section, and it consists of two open curves that extend outward. It might remind you of opposite-facing parabolas.

Key aspects of a hyperbola include:

• Two branches that are mirror images of each other.
• A center point from which the branches stretch away.
• Vertices located on the transverse axis, the line segment connecting points closest to each other on the branches.
• Foci that lie along the transverse axis, further away from the center than the vertices.
• Asymptotes that guide the hyperbola's open ends inward, shaping its curves.

The standard form of a hyperbola is:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]or\[\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1\]depending on its orientation.

Here,

• \(h, k\) define the center.
• \(a^2\) and \(b^2\) explain the relative distances of the axes.

From the equation, the parameters help identify the hyperbola's center, vertices, and foci. The asymptotes are calculated and sketched to help detail the hyperbola's curvature. A full understanding of hyperbolas enriches comprehension of mathematical models and architectures where these curves appear.