Let the common difference of the elements in the first AP and the second AP be \(h\) and \(k\), respectively.

Any line segment that connects two consecutive points has a slope equal to: \[ \frac{q_{i+1}-q_i}{p_{i+1}-p_i} = \frac{k}{h} \]

The points in \(S\) lie on the line that passes through \(\left(p_1,q_1\right)\) with slope \(k/h\). Therefore, all the points in \(S\) must be collinear.

The equation of the line is given by \(y=mx+2000\). Suppose the line touches the graph at \( (x_0,{x_0}^{3}) \). The slope of the tangent at that point is \(y^{\prime} = 3{x_0}^2 \). \begin{align} y &= mx + 2000 \\ {x_0}^3 &= (3{x_0}^2)x_0 + 2000 \\ x_0 &= -10 \\ y^{\prime} &= 3{x_0}^{2} = 300 \end{align} This implies that the slope of the line \(L\) is 300.

Center \(=(10,0),\) radius \(=6\). Using the distance formula and the Pythagorean theorem we get: \[y^{2}+(x+8)^{2}-6^{2}=9\left(y^{2}+(x-8)^{2}-4\right)\] Simplifying gives \(y^{2}+(x-10)^{2}=6^{2}\).

A bubble at the point \(P=\left(a, a^{2}\right)\) must be tangential to the parabola at \(\left(a, a^{2}\right)\), since the circle must lie within the region of the parabola. The circle must also be symmetric with respect to \(y\)-axis, since the parabola is symmetric. So its center \(O\) must be on the \(y\)-axis. See the figure below. The radius \(OP\) of this bubble is perpendicular to the common tangent to the parabola and to the bubble at \(P \). The slope of this tangent = \(2 a,\) so the slope of radius \(O P=\frac{-1}{2 a}(\) for \(a \neq 0) \).

Let \(Q=\left(0, a^{2}\right) \). Using the \(\Delta O P Q\), the slope of \(O P=\frac{-O Q}{a}=\frac{-1}{2 a}\). Therefore \(O Q=\frac{1}{2},\) regardless of the value of \(a\)

(a) We are given that the radius of the circle \(OP=1\) cm. By Pythagoras: \[ OP^{2}=\left(\frac{1}{2}\right)^{2}+a^{2}=1 \] So \(a^{2}=\frac{3}{4}\) and \(O=\left(0, \frac{3}{4}+\frac{1}{2}\right)=\left(0, \frac{5}{4}\right)\) Hence, the center of the bubble is at \( (0,5/4) \).

(b) For any nonzero \(a\), the radius of the bubble satisfies \(O P^{2}=\left(\frac{1}{2}\right)^{2}+a^{2},\) so \(O P>\frac{1}{2}\). The smallest bubble is at the origin and its radius is \(\frac{1}{2}\).

We are looking for a vector \( \vec{u} \) that can be written as \( t\vec{a}+\vec{b} \) for some \(t\). Since \( \vec{u} \) is perpendicular to \(\vec{c}\) we must have \( \vec{u}\cdot\vec{c} = 0 \).

The given property of the curve \(C\) can be expressed as a differential equation:

\[\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)\]

It is convenient to let \(z=y / x,\) so the equation becomes \(\frac{d y}{d x}=\frac{1}{2}\left(\frac{1}{z}+z\right).\)

To get this in terms of only \(x\) and \(z,\) differentiate \(z=y / x\) with respect to \(x\) to get: \[\frac{d z}{d x}=\frac{1}{x} \frac{d y}{d x}-\frac{y}{x^{2}}=\frac{1}{x}\left(\frac{d y}{d x}-z\right)=\frac{1}{x}\left(\frac{1}{2}\left(\frac{1}{z}+z\right)-z\right)=\frac{1}{x} \frac{1-z^{2}}{2 z}\]

Separating the variables and integrating, we get: \begin{align} \int \frac{d x}{x}&=\int \frac{2 z d z}{1-z^{2}}\\ \log |x|&=-\log \left|1-z^{2}\right|+ \text{some constant}\\ \log \left|1-z^{2}\right|&=-\log |x|+K=\log |x|^{-1}+K \\ 1-z^{2}&=\pm \frac{e^{K}}{x}=\frac{c}{x} \end{align}
where \(c\) is a nonzero constant. Substituting \(z=y / x,\) we get \(1-\frac{y^{2}}{x^{2}}=\frac{c}{x},\) i.e., \(x^{2}-y^{2}=c x\).

To get the shape of the curve, complete the square to get \(\left(x-\frac{c}{2}\right)^{2}-y^{2}=\frac{c^{2}}{4},\) which is a hyperbola when \(c \neq 0\). It could also represent two straight lines \(y=\pm x\), when \(c=0\).

Clearly the line \(L\) passes through the origin. Moreover \(L\) is in the direction perpendicular to the normals to the both the planes. The direction vector can be obtained by computing the following cross product: \[ (i+j) \times(j+\hat{k})=i-j+\hat{k} \] Hence \(L\) can be written as \(L=\{(0,0,0)+\lambda(1,-1,1) \mid \lambda\) is a real number \(\}\). First, note that the equation of any plane that contains the line \(L\) is given by \[ x+(1+\lambda) y+\lambda z=0 \] Second, note that one can rotate the plane \(x+y=0\) in either clockwise or in anticlockwise direction. Consequently there are two such planes. The normal of one of the planes makes an angle of \(45^{\circ}\) with the normal of \(x+y=0\) and the other normal makes an angle of \(135^{\circ}\). \[ \begin{array}{c} (i+j) \cdot(i+(1+\lambda) j+\lambda \hat{k})=\pm|i+j \| i+(1+\lambda) j+\lambda \hat{k}| \cos \left(\frac{\pi}{4}\right) \\ 2+\lambda=\pm \sqrt{1+(1+\lambda)^{2}+\lambda^{2}} \\ \lambda^{2}-2 \lambda-2=0 \\ \lambda=1 \pm \sqrt{3} \end{array} \] So the equation of the plane is \[ x+y+(1 \pm \sqrt{3})(y+z)=0 \]