Problem 41
Question
Find the standard form of the equation of the hyperbola with the given characteristics. Foci: (0,±8)\(;\) asymptotes: \(y=\pm 4 x\)
Step-by-Step Solution
Verified Answer
The standard form of the equation of the hyperbola with the given characteristics is \(y^2/16 - x^2/256 = 1\).
1Step 1: Calculate a and b based on given foci and asymptotes
The equation for foci is \(c=\sqrt{(a^2+b^2)}\) and for the asymptotes, in this case, we know b/a = 4. Since c is given as 8, and we know that c should be greater than a in case of hyperbola, we can assume a to be less than 8. By systematic trial and error, let's consider a=1, then b will be 4 (since b/a = 4). But this gives \(c=\sqrt{(1^2+4^2)}\) i.e., c = 4.169 which is not equal to 8. Proceed like this until we find that a= 4 gives b = 16 and hence \(\sqrt{16^2 + 4^2}\) = 8, So, a=4, b=16 and c= 8.
2Step 2: Write the Equation of the Hyperbola
We now have all the necessary components to write out the standard form of the hyperbola equation. The equation is \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\). Given that the center of the hyperbola is at origin (0,0), the constants h and k are both 0. Substituting a=4 and b=16 into the equation results in the final equation: \(y^2/16 - x^2/256 = 1\).
Key Concepts
Foci of a HyperbolaAsymptotes of a HyperbolaHyperbola Equation Derivation
Foci of a Hyperbola
When it comes to understanding hyperbolas, one of the most crucial elements is the concept of foci. The foci (singular: focus) are two fixed points located inside each of the hyperbola's two branches. The remarkable property of a hyperbola is that for any point on the curve, the absolute difference of the distances to the foci is constant.
A hyperbola's foci are important because they help define the shape and orientation of the curve. To find the foci of a hyperbola, we use the equation \(c = \sqrt{a^2 + b^2}\), where \(c\) is the distance from the center to a focus, \(a\) is the distance from the center to a vertex on the transverse axis, and \(b\) is the distance from the center to a vertex on the conjugate axis. In our considered problem, the distance to the foci is given as 8 units along the y-axis, and by using the relation between \(a\), \(b\), and \(c\), we can determine the values of \(a\) and \(b\) to construct the standard form of the hyperbola's equation.
A hyperbola's foci are important because they help define the shape and orientation of the curve. To find the foci of a hyperbola, we use the equation \(c = \sqrt{a^2 + b^2}\), where \(c\) is the distance from the center to a focus, \(a\) is the distance from the center to a vertex on the transverse axis, and \(b\) is the distance from the center to a vertex on the conjugate axis. In our considered problem, the distance to the foci is given as 8 units along the y-axis, and by using the relation between \(a\), \(b\), and \(c\), we can determine the values of \(a\) and \(b\) to construct the standard form of the hyperbola's equation.
Asymptotes of a Hyperbola
Another fundamental concept tied to hyperbolas is their asymptotes. Asymptotes are straight lines that the curve approaches but never actually reaches. These lines provide an envelope that dictates the spread of the hyperbola's branches.
Every hyperbola has two asymptotes, and they intersect at the hyperbola's center, creating an X-shape. The slopes of the asymptotes are determined by the equation \(y = \pm\frac{b}{a}x\) in the case of a hyperbola centered at the origin. Here \(a\) and \(b\) are the same values used to define the hyperbola's equation. In our exercise, the slopes are given directly as \(y=\pm 4x\), meaning the slope is 4 and -4. This indicates how steep the asymptotes are and subsequently guides us in sketching the hyperbola's general shape, providing a visual aid to understand the curve's spread in relation to its center.
Every hyperbola has two asymptotes, and they intersect at the hyperbola's center, creating an X-shape. The slopes of the asymptotes are determined by the equation \(y = \pm\frac{b}{a}x\) in the case of a hyperbola centered at the origin. Here \(a\) and \(b\) are the same values used to define the hyperbola's equation. In our exercise, the slopes are given directly as \(y=\pm 4x\), meaning the slope is 4 and -4. This indicates how steep the asymptotes are and subsequently guides us in sketching the hyperbola's general shape, providing a visual aid to understand the curve's spread in relation to its center.
Hyperbola Equation Derivation
Deriving the equation of a hyperbola in standard form is akin to assembling a puzzle - each piece of information helps reveal the bigger picture. For our problem, we start with the known values for the foci and asymptotes and use these to find the values of \(a\), \(b\), and \(c\), the key parameters of the hyperbola.
To illustrate, we know the hyperbola's foci are at (0,±8) and its asymptotes have the equations \(y=\pm 4x\). From the relationships \(c=\sqrt{a^2+b^2}\) and \(\frac{b}{a}=4\), we derived that \(a=4\) and \(b=16\). With these values, the standard form of a vertical hyperbola centered at the origin is \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\). Substituting \(a\), \(b\), \(h=0\), and \(k=0\) into this equation gives us the standard form \(y^2/16 - x^2/256 = 1\). This equation is the final piece we needed, completing the 'puzzle' and providing a rigorous mathematical description of the hyperbola in question.
To illustrate, we know the hyperbola's foci are at (0,±8) and its asymptotes have the equations \(y=\pm 4x\). From the relationships \(c=\sqrt{a^2+b^2}\) and \(\frac{b}{a}=4\), we derived that \(a=4\) and \(b=16\). With these values, the standard form of a vertical hyperbola centered at the origin is \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\). Substituting \(a\), \(b\), \(h=0\), and \(k=0\) into this equation gives us the standard form \(y^2/16 - x^2/256 = 1\). This equation is the final piece we needed, completing the 'puzzle' and providing a rigorous mathematical description of the hyperbola in question.
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Problem 41
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