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Chapter 11: Problem 33

Determine the equation of the hyperbola satisfying the given conditions. Write each answer in the form \(A x^{2}-B y^{2}=\) Cor in the form \(A y^{2}-B x^{2}=C\). Vertices (0,卤7)\(;\) graph passes through the point (1,9)

Short Answer

Expert verified
The equation of the hyperbola is \(32y^2 - 49x^2 = 49\).

Step by step solution

01

Identify the Hyperbola Type and the Equation Form

Since the vertices of the hyperbola are given as \((0, \pm 7)\), it indicates that the hyperbola is vertical. Therefore, the equation of the hyperbola is in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), where \(a\) is the distance from the center to a vertex along the y-axis.
02

Determine the Value of 'a'

From the vertices \((0, \pm 7)\), it is clear that the distance from the center (at origin) to a vertex is 7. Therefore, \(a = 7\) and hence \(a^2 = 49\). The equation of the hyperbola can now be partially written as \(\frac{y^2}{49} - \frac{x^2}{b^2} = 1\).
03

Use the Given Point to Find 'b^2'

The hyperbola passes through the point \((1, 9)\). Substitute \(x = 1\) and \(y = 9\) into the equation \(\frac{y^2}{49} - \frac{x^2}{b^2} = 1\) to find the value of \(b^2\).\[\frac{9^2}{49} - \frac{1^2}{b^2} = 1\]\[\frac{81}{49} - \frac{1}{b^2} = 1\]\[\frac{81}{49} - 1 = \frac{1}{b^2}\]\[\frac{32}{49} = \frac{1}{b^2}\]Thus, \(b^2 = \frac{49}{32}\).
04

Write the Equation in the Specified Form

Now we have the values needed to write the equation of the hyperbola. The equation derived so far is \(\frac{y^2}{49} - \frac{x^2}{\frac{49}{32}} = 1\). To convert this into the form \(Ay^2 - Bx^2 = C\), multiply every term by 49 to eliminate the fractions:\[32y^2 - 49x^2 = 49\]This is the equation in the desired form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertices of hyperbola
The term **vertices** refers to the points on a hyperbola that are closest to each other. These points dictate the shape and orientation of the hyperbola. In this case, the vertices are given as \((0, \pm 7)\), indicating a vertical hyperbola since these points lie along the y-axis.
This arrangement tells us that the center of the hyperbola is at the origin, \((0, 0)\). It implies that from this central point, the vertices extend 7 units both upwards and downwards.
Key points about vertices in hyperbolas:
  • The direction in which they lie indicates whether the hyperbola is vertical or horizontal.
  • The distance from the center to each vertex is denoted by \(a\).
  • The equation uses this distance to determine the parameter \(a^2\), essential for developing the hyperbola's equation.
Understanding vertices is crucial because it sets the framework for the rest of the hyperbola's geometry.
Equation of conic sections
Conic sections such as circles, ellipses, parabolas, and hyperbolas each have unique equations that describe their shapes. A hyperbola's equation is distinct because it consists of a difference between squares. Depending on whether it is vertical or horizontal, the equation takes different forms.
For a vertical hyperbola, as in this exercise, the equation starts as:
  • \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
Here:
  • \(a\) is the distance from the center to a vertex along the y-axis.
  • The \(x^2\) and \(y^2\) terms signify the shape and orientation, with their coefficients indicating how stretched or compressed the hyperbola is.
The conversion to the form \(Ay^2 - Bx^2 = C\) involves clearing denominators, showcasing how a vertical hyperbola diverges from the standard forms of other conics.
Distance from center to vertex in hyperbola
The **distance from the center to a vertex** is a fundamental concept when dealing with hyperbolas. This distance, represented by \(a\), determines key characteristics of the hyperbola.
Since the center for our hyperbola is at the origin, \((0, 0)\), and the vertices are known \((0, \pm 7)\), we find the distance is 7 units. This determines \(a = 7\). Therefore, \(a^2 = 49\).
This distance from the center to a vertex affects:
  • The formula of the hyperbola. The value of \(a^2\) appears in the equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
  • The overall size and openness of the hyperbola.
Exploring the distance from the center to vertex provides a deeper understanding of the hyperbola's structure and is key in developing its equation.

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Most popular questions from this chapter

Find the equation of the parabola satisfying the given conditions. In each case, assume that the vertex is at the origin. The focus lies on the \(y\) -axis, and the parabola passes through the point (7,-10)

A line is drawn tangent to the ellipse \(x^{2}+3 y^{2}=52\) at the point (2,4) on the ellipse. (a) Find the equation of this tangent line. (b) Find the area of the first-quadrant triangle bounded by the axes and this tangent line.

Let \(P\) be a point on the right-hand branch of the hyperbola \(x^{2}-y^{2}=k^{2} .\) If \(d\) denotes the distance from \(P\) to the center of the hyperbola, show that $$d^{2}=\left(F_{1} P\right)\left(F_{2} P\right)$$

Find the equation of the line that is tangent to the hyperbola at the given point. Write your answer in the form \(y=m x+b\). $$x^{2}-4 y^{2}=16 ;(5,3 / 2)$$

Consider the equation \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\) (a) Use the symmetry tests from Section 1.7 to explain why the graph of this equation must be symmetric about both the \(x\) -axis and the \(y\) -axis. (b) Show that solving the equation for \(y\) yields \(y=\pm \frac{1}{3} \sqrt{36-4 x^{2}}\) (c) Let \(y=\frac{1}{3} \sqrt{36-4 x^{2}} .\) Use the techniques of Section 2.4 to find the domain of this function. (You need to solve the inequality \(36-4 x^{2} \geq 0 .\) ) You should find that the domain is the closed interval [-3,3] (d) Use a calculator to complete the following table. Then plot the resulting points and connect them with a smooth curve. This gives you a sketch of the first quadrant portion of the graph of \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\) $$\begin{array}{llllllll}x & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 & 3 \\\\\hline y=\frac{1}{3} \sqrt{36-4 x^{2}} & 2 & & & & & & 0 \\\\\hline\end{array}$$ (e) Use your graph in part (d) along with the symmetry results in part (a) to sketch a complete graph of \(\left(x^{2} / 3^{2}\right)+\left(y^{2} / 2^{2}\right)=1\).
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