The General Second Degree Equation

Section 4.6 The General Second Degree Equation

By the end of this section, you should:

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know the general form of second degree equation representing conic sections whose lines of symmetry are not necessarily parallel to the coordinate axes.
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know the rotation formula for rotating the coordinate axes.
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be able to find equivalent equation of a conic section under rotation of the reference axes.
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be able to apply the rotation formula to find a suitable coordinate system in which a given general second degree equation is converted to a simpler standard form.
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be able to convert a given general second degree equation to an equivalent simpler standard form of equation of a conic section.
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be able to identify a conic section that a given general second degree equation represents and sketch the corresponding conic section.
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In the previous sections we have seen that, except in degenerate cases, the graph of the equation

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\(Ax_2 + Cy_2 + Dx + Ey + F = 0\)

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is a circle, parabola, ellipse or hyperbola. The construction of these equations was based on the assumption that the axis of symmetry of a conic section is parallel to one of the coordinate axes. The assumption seems to be quite restrictive because the axis of symmetry for a parabola, ellipse, or hyperbola can be any oblique line as indicated in their corresponding definitions.

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However, the reason why we have assumed that is not only for simplicity but there is always a coordinate system whose one of the axes is parallel to a desired line of symmetry. In particular, we can rotate the axes of our xy-coordinate system, whenever needed, so as to form a new x’y’-coordinate system such that either the x’- axis or y’-axis is parallel to the desired line of symmetry. Toward this end, let us review the notion of rotation of axes.

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Subsection 4.6.1 Rotation of Coordinate Axes

A rotation of the x and y coordinate axes by an angle θ about the origin O(0,0) creates a new x’y’-coordinate system whose x’-axis is the line obtained by rotating the x-axis by angle θ about O and y’-axis is the line obtained by rotating the y-axis in the same way. This makes a point P to have two sets of coordinates denoted by (x,y) and (x′,y′) relative to the xy- and x′y′-coordinate axes, respectively.(See Figure 4.6.1)

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The angle θ considered in the above discussion is called the angle of rotation. Our aim is to find the relationships between the coordinates (x,y) and the coordinates (x′,y′) of the same point P.

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To find this relationships, let P(x,y) be any point in xy-plane, θ be an angle of rotation (i.e., θ is angle between x and x’ axes ) and θ be the angle between OP and x’-axis. (See Figure 4.6.2)

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So, letting |OP| = r observe that

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\(x'=rcos\left ( \theta \right )\text{,}\) \(y'=rsin\left ( \theta \right )\) .....(1)

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and

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\(x'=rcos\left ( \theta +\phi \right )\text{,}\) \(y'=rsin\left ( \theta +\phi \right )\) .....(2)

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Then, using the trigonometric identities

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\(cos\left ( \theta +\phi \right )=cos\theta cos\phi -sin\theta sin\phi \)

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\(sin\left ( \theta +\phi \right )=sin\theta cos\phi -cos\theta sin\phi \)

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and (1), the equations in (2) become

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\(\begin{align} x & = x'cos\theta -y'sin\theta \\ y & = x'sin\theta -y'cos\theta .........(3) \end{align}\)

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Moreover, these equations can be solved for x′ and y′ in terms of x and y to obtain

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\(\begin{align} x' & = xcos\theta +ysin\theta \\ y' & = ysin\theta +ycos\theta ......(4) \end{align}\)

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The Equations (3) and (4) are called \(\textbf{rotation formulas}\text{.}\) It follows that if the angle of rotation θ is given, then Equation (3) can be used to determine the x and y coordinates of a point P if we know its x′ and y′ coordinates. Similarly, Equation (4) can be used to determine the x′ and y′ coordinates of P if we know its x and y coordinates.

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Example 4.6.3.

Suppose the x and y coordinate axes are rotated by π/4 about the origin.

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Find the coordinates of P(1, 2) relative to the new x’ and y’ axes.
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Find the equation of the curve xy = 1 relative to the new x’y’-coordinate system and sketch its graph
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Solution.

The given information about P and the curve are relative to the xy-coordinate system and we need to express them in terms of x’ and y’ coordinates relative to the new x′y′-coordinate system obtained under the rotation of the original axes by θ = π/4 rad about the origin. Thus, we use \(cos\frac{\pi}{4}=sin\frac{\pi}{4}= \frac{\sqrt{2}}{2}\) in the relevant rotation formula to obtain the following.

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Since P(1,2) has the coordinates x=1 and y=2, its x’ and y’ coordinates are, using formula (4)
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\begin{equation*} x'=\frac{\sqrt{2}}{2}\left ( 1 \right )+\frac{\sqrt{2}}{2}\left ( 2 \right )=\frac{3\sqrt{2}}{2} \end{equation*}

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\begin{equation*} y'=\frac{\sqrt{2}}{2}\left ( 1 \right )+\frac{\sqrt{2}}{2}\left ( 2 \right )=\frac{\sqrt{2}}{2} \end{equation*}

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Therefore, the coordinates of P relative to the new x’ and y’ axes are \((\frac{3\sqrt{2}}{2},\frac{\sqrt{2}}{2})\)
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We need to express x and y in the equation xy =1 in terms of x’ and y’ using the rotation formula (3). So, again since \(cos\left ( \frac{\pi}{4} \right )=sin\left ( \frac{\pi}{4} \right )=\frac{\sqrt{2}}{2}\) ,we obtain the formula (3)
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\begin{equation*} x=\frac{\sqrt{2}}{2}x'-\frac{\sqrt{2}}{2}y' \end{equation*}

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and

\begin{equation*} y=\frac{\sqrt{2}}{2}x'+\frac{\sqrt{2}}{2}y' \end{equation*}

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Therefore, xy=1
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\begin{equation*} \left(\frac{\sqrt{2}}{2}x'-\frac{\sqrt{2}}{2}y'\right )\left ( \frac{\sqrt{2}}{2}x'+\frac{\sqrt{2}}{2}y' \right )=1 \end{equation*}

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\begin{equation*} \left(\frac{\sqrt{2}}{2}x'\right )^{2}-\left ( \frac{\sqrt{2}}{2}y' \right )^{2}=1 \end{equation*}

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\begin{equation*} \frac{x'^{2}}{2}-\frac{y'^{2}}{2}=1 \end{equation*}

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Note that this is an equation of a hyperbola with center at origin vertices \(\left ( -\sqrt{2} ,0\right )\) and \(\left ( \sqrt{2} ,0\right )\) in the x’y’-coordinate system with principal axis on x’-axis. Since the x and y- axes were rotated though an angle of π/4 to obtain x’ and y’-axes, the hyperbola can be sketched as shown. (You may use Formula (3) to show that the vertices \(\left ( -\sqrt{2} ,0\right )\) and \(\left ( \sqrt{2} ,0\right )\) are (-1,1) and (1,1), respectively, relative to the x and y-axes).
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Example 4.6.5.

Find an equation of the ellipse whose center is the origin, vertices are (−4,−3) and (4, 3), and length of minor axis is 6.

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Solution.

The position of the ellipse is as shown in Figure 4.6.6 .

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To apply the standard equation of ellipse we use the x’y’-coordinate system such that the x’-axis coincide with the major axis of the ellipse. Therefore, the equation of the ellipse relative to the x’y’ system is

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\begin{equation*} \frac{x'^{2}}{a^{2}}=\frac{y'^{2}}{b'^{2}}=1 \end{equation*}

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Moreover, from the given information, \(a_2\) = \(|OV|_2\) =\(3_2\)+ \(4_2\) =25; and

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length of minor axis = 2b = 6 ⇒ b= 3. So, \(b_2\)= 9 .

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Hence, the equation of the ellipse relative to the x’y’ -coordinate system is

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\begin{equation*} \frac{x'^{2}}{25}=\frac{y'^{2}}{9}=1 \, \quad \text{or} \, \quad 9x'^{2}+25y'^{2}=225 ......(1) \end{equation*}

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Now we use the rotation formula to express the equation relative to our xy-coordinate system.

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So, let θ the angle between x-axis and x’-axis. Then, observe that

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\begin{equation*} cos\theta=\frac{4}{5} \, \quad \text{and} \, \quad sin\theta=\frac{3}{5} \end{equation*}

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Thus, using rotation formula (4) we get:

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\begin{equation*} x'=xcos\theta+ysin\theta = \frac{4}{5}x + \frac{3}{5}y \end{equation*}

\begin{equation*} y'=xsin\theta+ycos\theta = \frac{-3}{5}x + \frac{4}{5}y \end{equation*}

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Now we substitute these for x′ and y′ in (1) to obtain

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\begin{equation*} 9\left ( \frac{4}{5}x+\frac{3}{5}y \right )^{2}+25\left ( \frac{-3}{5}x+\frac{4}{5}y \right )^{2} =225 \end{equation*}

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And simplifying this we get

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\begin{equation*} 369x^2 - 384xy + 481y^2 - 5625 = 0 \end{equation*}

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which is the equation of the ellipse in the xy- coordinate system.

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Exercises Exercises

1.

Suppose the xy-coordinate axes are rotated 60° counterclockwise about the origin to obtain the new x’y’-coordinate system.

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If each of the following are coordinates of points relative to the xy-system, find the coordinates of the points relative to the x′y′- system.
1. (5, 0)
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2. (1, 4)
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3. (0, 1)
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4. (−1/2, 5/2)
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5. (−2, −1)
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Find the equation of the following lines and conics relative to the new x′y′- system.
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1. \(\displaystyle x = 5\)
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2. \(\displaystyle x − 2y = 1\)
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3. \(\displaystyle x^2 + y^2 = 1\)
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4. \(\displaystyle (x − 1)^2 + y^2 = 4\)
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5. \(\displaystyle x^2 − 4y = 1\)
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6. \(\displaystyle 4x^2 + (y − 2)^2 = 4\)
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7. \(\displaystyle x^2 + 4y^2 − 4x = 0\)
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8. \(\displaystyle x^2 − 4y^2 = 1\)
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9. \(\displaystyle −x^2 + y^2 − 2y = 0\)
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2.

Suppose the \(xy\)-coordinate axes are rotated 30° counterclockwise about the origin to obtain the new \(x'y'\)-coordinate system. If the following points are with respect to the new \(x'y'\)-system, what is the coordinates of each point with respect to the old \(xy\)-system?

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(0, 2)
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(−2, 4)
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(1, −3)
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\(\displaystyle \left ( \sqrt{3} ,-\sqrt{3}\right )\)
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Checkpoint 4.6.7.

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Checkpoint 4.6.8.

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Subsection 4.6.2 Analysis of the General Second Degree Equations

In the previous sections, we have seen that the equation of the form

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\(A x^2 + C y^2 + D x + E y + F = 0\text{..........}\) (1)

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represents a conic section (a parabola, ellipse, or hyperbola) whose axis of symmetry is parallel to one of the coordinate axes, except in degenerate cases.

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In Subsection 4.6.1, we also saw some examples of conic sections whose equations involve the \(xy\) term when their lines of symmetry are not parallel to either of the axes.

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Now we would like to analyze the graph of any quadratic (second degree) equation in \(x\) and \(y\) of the form:

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\(A x^2 + B x y + C y^2 + D x + E y + F = 0\text{...........}\) (2)

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where \(B ≠ 0\text{.}\) In order to analyze the graph of Equation (2), we usually convert it into an equation of type (1) using a suitable rotated coordinate system. To this end, we prove the following theorem.

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Theorem 4.6.9.

Consider the general second degree equation of the form (2), i..e.,

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\(A x^2 + B x y + C y^2 + D x + E y + F = 0\text{,}\) where \(B ≠ 0\) ........... (2)

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there is a rotation angle \(θ ∈ (0, π/2)\)through which the \(xy\)-coordinate system in which Equation (2) reduces to the form

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\(A' x'^2 + C' y'^2 + D' x' + E' y' + F' = 0\text{...........}\) (3)

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\(\textbf{Proof:}\) Let the \(xy\)-coordinate system be rotated by an angle \(θ\) about the origin to form a new \(x'y'\)-coordinate system. Then, from the rotation formula (3),we have

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\begin{equation*} x=x'cos\theta -y'sin\theta \,\, \text{and} \, \, y=x'sin\theta +y'cos\theta \end{equation*}

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we substitute for \(x\) and \(y\) in Equation (2) so that

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\begin{equation*} A\left ( x'cos\theta -y'sin\theta\right)^{2} +B\left ( x'cos\theta -y'cos\theta \right )\left ( x'sin\theta +y'cos\theta \right ) \end{equation*}

\begin{equation*} +C\left ( x'sin\theta +ycos'\theta \right )^{2}+ D\left ( x'cos\theta -y'sin\theta \right )+E\left ( x'sin\theta +y'cos\theta \right )+F=0 \end{equation*}

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After some calculations, combining like terms (those involving \(x'_2,x'y',y'_2\) and so on ), we get equation of the form

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\(A' x'^2 + B' x' y' + C' y'^2 + D' x' + E' y' + F' = 0\text{...............}\) (4)

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where the new coefficient \(B'\) is given by:

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\(B' = 2(C − A) sin(θ) cos(θ) + B (cos^2(θ) − sin^2(θ))\)

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Here the exact expressions for A’, C’, D’, E’ are omitted as they are irrelevant. What we need is to get the angle of rotation \(\theta\) for which Equation (4) has \(\textbf{no}\) x’y’ term, that is B’=0.

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This means that,

\begin{equation*} 2(C − A) sin(θ) cos(θ) + B (cos^2(θ) − sin^2(θ))=0 \end{equation*}

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Since \(2 sin(θ) cos(θ) = sin2(θ)\) and \(cos^2(θ) + B − sin^2(θ)=cos2θ\)

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this equation is equivalent to

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\begin{equation*} (C − A) sin(2θ) + B cos(2θ) = 0 \end{equation*}

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\begin{equation*} \text{or} \, \quad \frac{\text{cos}2\theta}{\text{sin}2\theta}=\frac{A-C}{B}, \text{Since} B \neq 0 \end{equation*}

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\begin{equation*} \text{or} \, \quad \text{cot}2\theta=\frac{A-C}{B} .......(5) \end{equation*}

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That is, if we choose the angle of rotation \(\theta\) satisfying (5), then B’ = 0 in Equation (4) so that the resulting equation in x’y’-coordinate system is in the form of Equation (3). Moreover, we can always find an angle that satisfies \(\text{cot}2\theta=\frac{A-C}{B}\) for any A C, \(\text{B} \, \epsilon \, \Re , \text{B}\neq 0\) since the range of the cotangent function is the entire set of real numbers. Note also that since \(2 \,\,\theta \, \epsilon (0,\pi )\text{,}\) the angel of rotation  can always be chosen so that \(0< \theta < \frac{\pi}{2}\text{.}\) So, the Theorem is proved.

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Remark 4.6.10.

If \(A = C\text{,}\) then \(\frac{\cos 2\theta}{\sin 2\theta} = \frac{A - C}{B} = 0 \Rightarrow 2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4}\text{.}\)

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Therefore, we can rewrite the result of the above theorem as follows:

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The rotation of the \(xy\)-coordinate system by angle \(\theta\) creates an \(x'y'\)-coordinate system in which a general second degree equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\text{,}\) with \(B \ne 0\text{,}\) is converted to an equation \(A'x'^2 + C'y'^2 + D'x' + E'y' + F' = 0\) (with no \(x'y'\) term), if we choose \(\theta \in (0, \frac{\pi}{2})\) such that:

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\begin{equation*} \tan 2\theta = \frac{B}{A - C}, \text{ if } A \ne C \end{equation*}

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\begin{equation*} \theta = \frac{\pi}{4}, \text{ if } \text{A = C} \end{equation*}

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Example 4.6.11.

Use rotation of axes to eliminate the \(xy\) term in each of the following equations,\ describe the locus (type of conic section) and sketch the graph of the equation

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\(\displaystyle 8x^2 + 28xy + 8y^2 - 32x - 32y + 30 = 0\)
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\(\displaystyle 73x^2 - 72xy + 52y^2 + 30x + 40y - 75 = 0\)
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Solution.

Given:
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\(8x^2 + 28xy + 8y^2 - 32x - 32y + 30 = 0 \Rightarrow A = C = 8\text{.}\) So, from the above Remark, the rotation angle is \(\theta = \frac{\pi}{4}\) \(\Rightarrow \cos\theta = \sin\theta = \frac{1}{\sqrt{2}}\text{,}\) \(\Rightarrow x = \frac{x' - y'}{\sqrt{2}}\) and \(y = \frac{x' + y'}{\sqrt{2}}\text{.}\)
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Now we substitute these for \(x\) and \(y\) in the given equation:
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\(8\left( \frac{x' - y'}{\sqrt{2}} \right)^2 + 28\left( \frac{x' - y'}{\sqrt{2}} \right)\left( \frac{x' + y'}{\sqrt{2}} \right)\)
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\(+ 8\left( \frac{x' + y'}{\sqrt{2}} \right)^2 - 32\left( \frac{x' - y'}{\sqrt{2}} \right) - 32\left( \frac{x' + y'}{\sqrt{2}} \right) + 30 = 0\)
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Expanding the squared expressions, combining like terms and simplifying, we obtain
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\(8x'^2 + 16y' - 32 = 0\) or, \(8(x'^2 + 2y' - 4) = 0\)
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This is an equation of a parabola. Its vertex is \((h', k') = (0, 2)\) relative to the \(x'y'\)-system, principal axis is on \(y'\)-axis and opens towards negative \(y'\) direction. (You can show that its vertex is \((h, k) = \left( \frac{2}{\sqrt{2}}, -\frac{2}{\sqrt{2}} \right)\) relative to the \(xy\)-system.) The graph of the equation is sketched in Figure 4.6.12
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Figure 4.6.12.
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Given: \(73x^2 - 72xy + 52y^2 + 30x + 40y - 75 = 0 \Rightarrow A = 73\text{,}\) \(B = -72\) and \(C = 52\text{.}\)
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Hence,
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\(\tan 2\theta = \frac{A - C}{B} = \frac{21}{-72} = -\frac{7}{24}\)
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The terminal side of \(2\theta\) is through \((-7, 24)\) since \(0 < 2\theta < \pi\text{.}\)
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\(\cos 2\theta = -\frac{7}{25}\text{.}\) Now as \(0 < \theta < \frac{\pi}{2}\text{,}\) both \(\cos\theta\) and \(\sin\theta\) are positive. Hence,
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\(\cos\theta = \sqrt{\frac{1 + \cos 2\theta}{2}} = \sqrt{\frac{1 - 7/25}{2}} = \frac{3}{5}\)
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\(\sin\theta = \sqrt{\frac{1 - \cos 2\theta}{2}} = \sqrt{\frac{1 + 7/25}{2}} = \frac{4}{5}\)
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This implies the \(x'\)-axis is through the coordinate point \((3, 4)\text{,}\) that is the line \(y = \frac{4}{3}x\text{.}\)
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Therefore, using the rotation formula (3), we get
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\(x = \frac{3}{5}x' - \frac{4}{5}y'\) and \(y = \frac{4}{5}x' + \frac{3}{5}y'\)
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Now we substitute these for \(x\) and \(y\) in the given equation to obtain
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\(73\left( \frac{3x' - 4y'}{5} \right)^2 - 72\left( \frac{3x' - 4y'}{5} \right)\left( \frac{4x' + 3y'}{5} \right) + 52\left( \frac{4x' + 3y'}{5} \right)^2 \)
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\(+ 30\left( \frac{3x' - 4y'}{5} \right) + 40\left( \frac{4x' + 3y'}{5} \right) - 75 = 0\)
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Expanding the squared expressions, combining like terms and simplifying, we obtain
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\(25x'^2 + 100x' - 50y' + 75 = 0\)
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Completing the square and dividing by 100 to get \(\frac{(x' + 2)^2}{4} + \frac{y'}{1} = 1\)
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which is an ellipse with center at \((h',k') = (-1, 0)\) relative to the \(x'y'\)-system, major axis on \(x'\)- axis (which is the line \(y = \frac{4}{3}x\) ), length of major axis = 4 and length of minor axis = 2. (You can show that the center is \((h, k) =\left( -\frac{3}{5}, -\frac{4}{5} \right)\) relative to the \(xy\)-system). The graph of the equation is sketched in Figure 4.6.13
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Figure 4.6.13.
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Exercises Exercises

1.

Find an equation of the conic section having the given properties and sketch its graph.

Ellipsi with center at origin, foci at \((-2,2)\) and \((2, 2)\text{,}\) and length of major axis \(2\sqrt{8}\text{.}\)
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Parabola whose vertex is at \((3, 4)\) and focus \((-5,-2)\)
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Hyperbola whose foci are \((-2, 2)\) and \((2, -2)\text{,}\) and length of transverse axis \(2\sqrt{2}\text{.}\)
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2.

Use rotation of axes to eliminate the \(xy\) term in each of the following equations, describe the locus (type of conic section) and sketch the graph of the equation.

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\(\displaystyle 17x^2 - 12xy + 8y^2 - 36 = 0\)
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\(\displaystyle 8x^2 + 24xy + y^2 - 1 = 0\)
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\(\displaystyle x^2 - 2xy + y^2 - 5y = 0\)
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\(\displaystyle 2x^2 + xy = 0\)
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\(\displaystyle 5x^2 + 6xy + 5y^2 - 4x + 4y - 4 = 0\)
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\(\displaystyle x^2 + 4xy + 4y^2 + 2x - 2y + 1 = 0\)
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3.

Show that if \(B > 0\text{,}\) then the graph of

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\begin{equation*} x^2 + 2Bxy = F \end{equation*}

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is a hyperbola if \(F \ne 0\text{,}\) and two intersecting lines if \(F = 0\)

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Checkpoint 4.6.14.

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Checkpoint 4.6.15.

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