Problem 9
Question
Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$x^{2}-y^{2}=1$$
Step-by-Step Solution
Verified Answer
Vertices: (1, 0), (-1, 0); Foci: (√2, 0), (-√2, 0); Asymptotes: y=x, y=-x. Graph: Two branches on x-axis.
1Step 1: Identify the Standard Form
The given equation of the hyperbola is \( x^{2} - y^{2} = 1 \). We can recognize this as a hyperbola in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) where \( a^2 = 1 \) and \( b^2 = 1 \). Thus, the values of \( a \) and \( b \) are both 1.
2Step 2: Find the Vertices
The vertices of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are located \( a \) units away from the center along the x-axis since it is oriented along x. So, the vertices are at \( (\pm a, 0) = (\pm 1, 0) \). Thus, the vertices are \( (1, 0) \) and \( (-1, 0) \).
3Step 3: Find the Foci
The foci of the hyperbola are calculated using \( c^2 = a^2 + b^2 \). Since \( a^2 = 1 \) and \( b^2 = 1 \), we get \( c^2 = 1 + 1 = 2 \). Thus, \( c = \sqrt{2} \). The foci are \( (\pm \sqrt{2}, 0) \) which are \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \).
4Step 4: Determine the Asymptotes
The equations of the asymptotes for the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by \( y = \pm \frac{b}{a}x \). Here \( a = 1 \) and \( b = 1 \), so the equations simplify to \( y = \pm x \). Therefore, the asymptotes are \( y = x \) and \( y = -x \).
5Step 5: Sketch the Hyperbola
Begin by plotting the vertices at \( (1, 0) \) and \( (-1, 0) \), and the foci at \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \). Draw the asymptotes as straight lines with equations \( y = x \) and \( y = -x \). The hyperbola approaches these lines as it extends outward from the vertices. Sketch the curve by drawing the two distinct branches approaching the asymptotes through the vertices.
Key Concepts
Vertices of a HyperbolaFoci of a HyperbolaAsymptotes of a Hyperbola
Vertices of a Hyperbola
When understanding hyperbolas, it's crucial to start with their vertices as these define the shape and orientation. The vertices of a hyperbola are specific points where the hyperbola intersects its transverse axis. In the standard form of a hyperbola, \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \] the orientation aligns along the x-axis, meaning the vertices are horizontally placed. Here, the value of \(a\) is crucial, and represents the distance from the center to each vertex along the x-axis. Given that both \(a^2 = 1\) and \(b^2 = 1\), the vertices for our particular equation \[x^2 - y^2 = 1 \] are found at
- \((1, 0)\)
- \((-1, 0)\)
Foci of a Hyperbola
Foci are vital in understanding the structure of a hyperbola. They are two fixed points located along the transverse axis that define the locus of points forming the hyperbola. The relationship between the center, vertices, and foci forms the distinctive structure that differentiates hyperbolas from other conic sections. To find the foci, we utilize the formula:\[ c^2 = a^2 + b^2 \] where \(c\) represents the distance from the center to each focus. For the equation \[x^2 - y^2 = 1,\] we calculate
- \(a^2 = 1\)
- \(b^2 = 1\)
- \(c^2 = 1 + 1 = 2\)
- \(c = \sqrt{2}\)
- \((\sqrt{2}, 0)\)
- \((-\sqrt{2}, 0)\)
Asymptotes of a Hyperbola
Asymptotes are straight lines that the branches of a hyperbola approach but never intersect. They give a sense of direction and boundary, without actually being a part of the hyperbola itself. These imaginary lines are essential in graphing the hyperbola accurately. For the standard form \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \] the equations of the asymptotes can be derived from:
- \(y = \pm \frac{b}{a}x\)
- \(y = x\)
- \(y = -x\)
Other exercises in this chapter
Problem 9
Determine the equation of the given conic in \(X Y\) -coordinates when the coordinate axes are rotated through the indicated angle. $$x^{2}-y^{2}=2 y, \quad \ph
View solution Problem 9
Find the center, foci, vertices, and asymptotes of the hyperbola. Then sketch the graph. $$\frac{(x+1)^{2}}{9}-\frac{(y-3)^{2}}{16}=1$$
View solution Problem 9
Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$x^{2}+4 y^{2}=16$$
View solution Problem 9
Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$x^{2}=9 y$$
View solution