Ch. 11 - Parametric Equations and Polar Coordinates

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 11 - Parametric Equations and Polar Coordinates

Problem 11.6.53

Chapter 11, Problem 11.6.53

Shifting Conic Sections

You may wish to review Section 1.2 before solving Exercises 39-56.

Exercises 53-56 give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes.

x²/4 − y²/5 = 1, right 2, up 2

Verified step by step guidance

Identify the original hyperbola equation: \(\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1\). This is a hyperbola centered at the origin \((0,0)\) with the transverse axis along the x-axis.

Apply the given shifts: right 2 units and up 2 units. This means replacing \(x\) by \((x - 2)\) and \(y\) by \((y - 2)\) in the equation to shift the center to \((2, 2)\).

Write the new equation of the hyperbola after shifting: \(\frac{(x - 2)^{2}}{4} - \frac{(y - 2)^{2}}{5} = 1\).

Determine the new center from the shifts: the center is now at \((2, 2)\).

Find the new vertices, foci, and asymptotes by using the standard formulas for hyperbolas centered at \((h, k)\): - Vertices: \((h \pm a, k)\) where \(a^{2} = 4\). - Foci: \((h \pm c, k)\) where \(c^{2} = a^{2} + b^{2} = 4 + 5 = 9\). - Asymptotes: lines through the center with slopes \(\pm \frac{b}{a} = \pm \frac{\sqrt{5}}{2}\), so equations are \(y - k = \pm \frac{\sqrt{5}}{2}(x - h)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Standard Form of a Hyperbola

A hyperbola's equation in standard form is typically written as (x - h)²/a² - (y - k)²/b² = 1 or its vertical counterpart. Here, (h, k) is the center, and a and b determine the distances to vertices and asymptotes. Understanding this form helps identify key features like center, vertices, foci, and asymptotes.

Translation (Shifting) of Conic Sections

Shifting a conic involves replacing x with (x - h) and y with (y - k) to move the graph h units right/left and k units up/down. This changes the center of the conic without altering its shape. Applying translations correctly updates the equation and the coordinates of key points.

Finding Foci and Asymptotes of a Hyperbola

The foci of a hyperbola lie along the transverse axis, found using c² = a² + b², where c is the focal distance from the center. Asymptotes are lines that the hyperbola approaches, with slopes ±b/a (horizontal transverse axis) or ±a/b (vertical). Calculating these requires knowing a, b, and the center.

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Related Practice

Textbook Question

Tangent Lines to Parametrized Curves

In Exercises 1−14, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d²y/dx² at this point.

x = sec² t − 1, y = tan t, t = −π/4

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Textbook Question

Ellipses

Exercises 25 and 26 give information about the foci and vertices of ellipses centered at the origin of the xy−plane. In each case, find the ellipse's standard−form equation from the given information.

Foci: ( ±√2, 0) Vertices: (±2,0)

Textbook Question

Parabolas

Exercises 9-16 give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Include the focus and directrix in your sketch.

x = −3y²

Textbook Question

Parabolas

Exercises 9-16 give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Include the focus and directrix in your sketch.

x² = 6y

Textbook Question

Hyperbolas

Exercises 27-34 give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch.

8x² − 2y² = 16

Textbook Question

Polar to Cartesian Equations

Replace the polar equations in Exercises 27–52 with equivalent Cartesian equations. Then describe or identify the graph.

r = 3 cos θ

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