Exercise 1: The Equations for the Circle and the Sphere in Standard Form
In an ordinary $(O,\mathbf i, \mathbf j)$-coordinate system in the plane a circle can – as is well known – be described by a quadratic equation in standard form
$$\,(x-c_1)^2+(y-c_2)^2=r^2\,$$
where $(c_1,c_2)$ is the centre and $r$ is the radius.
A
State the standard equation for the circle in the figure.
answer
$\,(x-3)^2+(y+1)^2=4\,.$
B
A circle has the equation
$$\,x^2+y^2+8x-6y=0\,.$$
Put it in standard form and by this determine the centre and the radius.
answer
Center =$(-4,3)$. Radius=5.
C
A spehre in $(x,y,z)$-space has the equation
$$x^2+y^2+z^2-2x+4y-6z+13 = 0\,.$$
Put it in the standard form
$$(x-c_1)^2+(y-c_2)^2+(z-c_3)^2=r^2$$
and by this determine the centre and the radius.
answer
Center =$(1,-2,3)$. Radius=1.
Exercise 2: The Standard Equation for the Three Typical Conic Sections
Intro: In the following examples we look at quadratic equations in more variables without mixed terms. Here it is possible to proceed by removing the linear terms. This technique is called completing the square and was practiced in Exercise 1 with circles and sphere. In the following we shall use the technique as a step towards the identification of conic sections.
A
An ellipse in the $(x,y)$-plane centered at $(c_1,c_2),$ and with the semi-axes $a$ and $b$ and the axes of symmetry $x=c_1$ and $y=c_2$ has the standard equation
Complete the square, put the equation in standard form, and state the centre of the ellipse, the semi-axes and the axes of symmetry.
hint
Plot the given equation using the Maple command implicitplot and check your results.
answer
Centre is $(-1,3).$ Semi-axes $a=1,\,b=2$. Axes of symmetry $x=-1,\,y=3.$
B
A hyperbola in the $(x,y)$-plane centered at $(c_1,c_2),$ with the semi-axes $a$ and $b$ the axes of symmetry $x=c_1$ and $y=c_2$ has the standard equation
Complete the square, put the equation in standard form, and state the centre of the hyperbola, the semi-axes and the axes of symmetry.
hint
Plot the given equation using the Maple command implicitplot and check your results.
answer
Centre is $(2,-2).$ Semi-axes $a=2,\,b=2$. Axes of symmetry $x=2,\,y=-2.$
C
A parabola in the $(x,y)$-plane with vertex $(c_1,c_2)$ and axis of symmetry $x=c_1$ has the standard equation
$$y-c_2=a(x-c_1)^2.$$
Or alternatively, if the parabola is not vertical but horizontal, making the axis of symmetry $y=c_2$:
$$x-c_1=a(y-c_2)^2.$$
A parabola is given by the equation
$$\,2x^2+12x-y+17=0\,.$$
Complete the square, put the equation in standard form, and state the vertex of the parabola and the axis of symmetry.
hint
Plot the given equation using the Maple command implicitplot and check your results.
answer
Vertex $(-3,-1).$ Axis of symmetry $x=-3.$
Exercise 3: Identification of a Conic Section
In an ordinary $(O,\mathbf i, \mathbf j)$-coordinate system in the plane a conic section is given by the following quadratic equation in two vaiables
$$9x^2+16y^2-24xy-40x-30y+250=0.$$
We shall find the type an characteristics of the conic section and proceed step by step.
A
The left-hand side of the quadratic equation can be split into two terms, the first terme is a quadratic form, – let’s call this $k(x,y),$ – and the second term is a linear polynomial. State the quadratic form and find its Hessian matrix.
answer
$k(x,y)=9x^2+16y^2-24xy\,.$
The Hessian matrix $\displaystyle{\mathbf H=\begin{matr}{rr}18&-24\\-24&32\end{matr}\,.}$
B
Determine a symmetric $2\times 2$-matrix $\mA$ that fulfills
$$\,\displaystyle{k(x,y)=\begin {matr}{cc} x & y \end{matr}\mA\begin{matr}{rr} x \\\\ y \end{matr}\,,}$$
and find a positive orthogonal matrix $\mathbf{Q}$ and a diagonal matrix $\mathbf{\Lambda}$, such that
$\mA$ you may read directly from the expression for $k?$ Otherewise you can get it from multiplying $\mathbf H$ by $\frac 12.$
answer
There are more options, here we choose
$\displaystyle{\mathbf Q=\begin{matr}{rr}\frac{3}{5}&\frac{4}{5}\\-\frac{4}{5}&\frac{3}{5}\end{matr}\quad\mathrm{and}\quad \Lambda= \begin{matr}{rr}25&0\\0&0\end{matr}}\,.$
C
Now state the form that $k$ attain after the reduction.
answer
Here we shall need the eigenvalues! If we call the new variables $x_1$ and $y_1,$ we get
$k(x_1,y_1)=25\cdot x_1^{\,2}+0\cdot y_1^{\,2}=25\cdot x_1^{\,2}\,.$
D
In a new coordinate system $(0,\mathbf q_1,\mathbf q_2),$ that appears by rotating$(O,\mathbf i, \mathbf j),$ the equation for the given conic section is changed so as to be without mixed terms. State the orthonormal basis that is part of the new coordinate system, and determine the new equation for the conic section.
hint
Of course, we use the columns of $\mQ$ as the new basis.
We have already changed the quadratic form to the new $(x_1,y_1)$-coordinates. So we only lack the linear terms $-40x-30y\,.$
hint
We have the change of basis relation $\displaystyle{\begin{matr}{r}x\\y\end{matr}=\mathbf Q\,\begin{matr}{r}x_1\\y_1\end{matr}}.$ From this we see how $x$ and $y$, respectively, can be expressed by $x_1$ and $y_1$ which one inserts in $-40x-30y\,.$
answer
$\displaystyle{
(\mathbf q_1,\mathbf q_2)=\left(\begin{matr}{r}\frac{3}{5}\\-\frac{4}{5}\end{matr},\begin{matr}{r}\frac{4}{5}\\\frac{3}{5}\end{matr}\right)}.$
The conic section has in the new coordinate system the equation $25x_1^{\,2}-50y_1+250=0\,.$
E
Which type of conic section are we talking about? State its characteristics, both in the new and in the old coordinate system. Illustrate using Maple.
answer
The equation for the conic section can immediately be rewritten as $\,\displaystyle{y_1-5=\frac 12\,x_1^{\,2}\,.}$ it is a parabola with vertex $(x_1,y_1)=(0,5)$ and the $\,y_1$-axis as axis of symmetry.
In the given coordinate system the vertex is given by $\displaystyle {\begin{matr}{r}x\\y\end{matr}=\mathbf Q\,\begin{matr}{r}0\\5\end{matr}=\begin{matr}{r}4\\3\end{matr}}.$
A directional vector for axis of symmetry in the new coordinate system is $\mathbf e=(0,1).$ In the old coordinates $\displaystyle{ \mathbf Q\,\begin{matr}{r}0\\1\end{matr}=\begin{matr}{r}\frac 45\\\frac 35\end{matr}}.$ Thus the axis of symmetry has the parametric representation
In an ordinary orthogonal $(x,y)$-coordinate system in the plane a curve is given by the equation:
$$52x^2+73y^2-72xy-200x-150y+525=0.$$
Describe the type and position of the curve, and state parametric representations for possible axes of symmetry.
hint
State the quadratic form in matrix form $\,\begin {matr}{cc} x & y \end{matr}\mA\begin{matr}{rr} x \\ y \end{matr}\,$ and diagonalize $\,\mA\,.$
hint
State the equation that the conic section has in the new $(x_1,y_1)$-coordinate system, that consists of the origin and the new orthonomal basis. First we deal with the quadratic form, then the linear terms.
hint
The conic section has in the new $(x_1,y_1)$-coordinate system the equation
$$100x_1^{\,2}+25y_1^{\,2}-250y_1+525=0\,.$$
now we complete the square after which the conic section is identified and characterized!
answer
In the new coordinate system the conic section has the equation
$$\frac {x_1^{\,2}}1+\frac{(y_1-5)^2}{4}-1=0$$
or similar (depending on your choice of a new basis). So, it is an ellipse with semi-axes $a=1$ and $b=2$ and center in $(0,5).$
In the given coordinate system the center is $(4,3),$ and the axes of symmetry are described by the parametric representations:
Major axis: $\,\displaystyle{(x,y)=(4,3)+t_1\big(\frac 45,\frac 35\big)}\,$ where $\,t_1\in\reel$
Minor axis: $\,\displaystyle{(x,y)=(4,3)+t_2\big(-\frac 35,\frac 45\big)}\,$ where $\,t_2\in\reel$
B
Plot the conic sections and the axis of symmetry using Maple and compare with a plot of level curves for the polynomial
$$f(x,y)=52x^2+73y^2-72xy-200x-150y+525\,.$$
In particular, consider the level curve corresponding to the height 0!
