8.4 Rotation of axes (Page 2/8)

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Identifying a conic from its general form

Identify the graph of each of the following nondegenerate conic sections.

$4 x^{2} - 9 y^{2} + 36 x + 36 y - 125 = 0$
$9 y^{2} + 16 x + 36 y - 10 = 0$
$3 x^{2} + 3 y^{2} - 2 x - 6 y - 4 = 0$
$- 25 x^{2} - 4 y^{2} + 100 x + 16 y + 20 = 0$

Rewriting the general form, we have
$A = 4$ and $C = −9,$ so we observe that $A$ and $C$ have opposite signs. The graph of this equation is a hyperbola.
Rewriting the general form, we have
$A = 0$ and $C = 9.$ We can determine that the equation is a parabola, since $A$ is zero.
Rewriting the general form, we have
$A = 3$ and $C = 3.$ Because $A = C,$ the graph of this equation is a circle.
Rewriting the general form, we have
$A = −25$ and $C = −4.$ Because $A C > 0$ and $A \neq C,$ the graph of this equation is an ellipse.

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Try It

Identify the graph of each of the following nondegenerate conic sections.

$16 y^{2} - x^{2} + x - 4 y - 9 = 0$
$16 x^{2} + 4 y^{2} + 16 x + 49 y - 81 = 0$

hyperbola
ellipse

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Finding a new representation of the given equation after rotating through a given angle

Until now, we have looked at equations of conic sections without an $x y$ term, which aligns the graphs with the x - and y -axes. When we add an $x y$ term, we are rotating the conic about the origin. If the x - and y -axes are rotated through an angle, say $θ,$ then every point on the plane may be thought of as having two representations: $(x, y)$ on the Cartesian plane with the original x -axis and y -axis, and $(x^{'}, y^{'})$ on the new plane defined by the new, rotated axes, called the x' -axis and y' -axis. See [link] .

The graph of the rotated ellipse $x^{2} + y^{2} - x y - 15 = 0$

We will find the relationships between $x$ and $y$ on the Cartesian plane with $x^{'}$ and $y^{'}$ on the new rotated plane. See [link] .

The Cartesian plane with x - and y -axes and the resulting x ′− and y ′−axes formed by a rotation by an angle $θ .$

The original coordinate x - and y -axes have unit vectors $i$ and $j .$ The rotated coordinate axes have unit vectors $i^{'}$ and $j^{'} .$ The angle $θ$ is known as the angle of rotation . See [link] . We may write the new unit vectors in terms of the original ones.

$\begin{array}{l} i^{'} = \cos θ i + \sin θ j \\ j^{'} = - \sin θ i + \cos θ j \end{array}$

Relationship between the old and new coordinate planes.

Consider a vector $u$ in the new coordinate plane. It may be represented in terms of its coordinate axes.

$\begin{array}{l} u = x^{'} i^{'} + y^{'} j^{'} \\ u = x^{'} (i \cos θ + j \sin θ) + y^{'} (- i \sin θ + j \cos θ) & \begin{matrix} \end{matrix} Substitute . \\ u = i x' \cos θ + j x' \sin θ - i y' \sin θ + j y' \cos θ & \begin{matrix} \end{matrix} Distribute . \\ u = i x' \cos θ - i y' \sin θ + j x' \sin θ + j y' \cos θ & \begin{matrix} \end{matrix} Apply commutative property . \\ u = (x' \cos θ - y' \sin θ) i + (x' \sin θ + y' \cos θ) j & \begin{matrix} \end{matrix} Factor by grouping . \end{array}$

Because $u = x^{'} i^{'} + y^{'} j^{'},$ we have representations of $x$ and $y$ in terms of the new coordinate system.

$\begin{matrix} x = x^{'} \cos θ - y^{'} \sin θ \\ and \\ y = x^{'} \sin θ + y^{'} \cos θ \end{matrix}$

A General Note

Equations of rotation

If a point $(x, y)$ on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle $θ$ from the positive x -axis, then the coordinates of the point with respect to the new axes are $(x^{'}, y^{'}) .$ We can use the following equations of rotation to define the relationship between $(x, y)$ and $(x^{'}, y^{'}) :$

$x = x^{'} \cos θ - y^{'} \sin θ$

and

$y = x^{'} \sin θ + y^{'} \cos θ$

How To

Given the equation of a conic, find a new representation after rotating through an angle.

Find $x$ and $y$ where $x = x^{'} \cos θ - y^{'} \sin θ$ and $y = x^{'} \sin θ + y^{'} \cos θ .$
Substitute the expression for $x$ and $y$ into in the given equation, then simplify.
Write the equations with $x^{'}$ and $y^{'}$ in standard form.

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Source: OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3

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