Expert Help with High School Geometry

Distance between (4,5) and (-6,-7)?

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$
Find the maximum area of an isosceles triangle inscribed in the ellipse $x^2/a^2+y^2/b^2=1$. My teacher solved it by considering two arbitrary points on the ellipse to be vertices of the triangle, being $(a\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},b\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$ and $(a\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},-<!--\; -\; -->b\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})$. (Let's just say $\mathrm{\theta <!--\; \theta \; -->}$ is theta) and then proceeded with the derivative tests(which i understood) But, he didn't indicate what our $\mathrm{\theta <!--\; \theta \; -->}$ was,and declared that these points always lie on an ellipse. Why so? And even if they do, what's the guarantee that points of such a form will be our required vertices? One more thing, I'd appreciate it if you could suggest another way of solving this problem.

Finding the volume of the tetrahedron with vertices (0,0,0), (2,0,0), (0,2,0), (0,0,2). I get 8; answer is 4/3.
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors A, B and C then the area of the object is $A\cdot <!--\; \cdot \; -->(B\times <!--\; \times \; -->C)$. Let V be the volume we are trying to find.
$\begin{array}{rl}{x}^2& =6-<!--\; -\; -->y^2-<!--\; -\; -->z^2\\ A& =(2,0,0)-<!--\; -\; -->(0,0,0)=(2,0,0)\\ B& =(0,2,0)-<!--\; -\; -->(0,0,0)=(0,2,0)\\ C& =(0,0,2)-<!--\; -\; -->(0,0,0)=(0,0,2)\\ V& =\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=\left|\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right|\\ & =2\left|\begin{array}{cc}2& 0\\ 0& 2\end{array}\right|=2(4-<!--\; -\; -->0)\\ & =8\end{array}$
However, the book gets $\frac{4}{3}$

Geometric Distribution - transformation $Y=x/(x+1)$
I honestly have no idea where to start with this problem:
Suppose X has the geometric distribution with PMF
$f(x)=\frac{1}{3}{\left(\frac{2}{3}\right)}^x,x=1,2,.....$
Determine the probability distribution of $Y=x/(x+1).$ Note that both X and Y are discrete random variables. To specify the probability distribution of Y, specify its PMF.

Cumulative geometric probability (greater than a value)
Emelia registers vehicles for the Department of Transportation. Sports utility vehicles (SUVs) make up 12% of the vehicles she registers. Let V be the number of vehicles Emelia registers in a day until she first registers an SUV. Assume the type of each vehicle is independent. Find the probability that Emelia registers more than 4 vehicles before she registers an SUV.

Distance between (-2,1,-3) and (5,6,-2)?

Distance between (8,6,2) and (0,6,0)?

Distance between (-8,17) and (-11,33)?

A different approach to a common question
A unit stick is randomly broken into 3 pieces, it is given that these three pieces can make a triangle, what is the expected length of the medium-sized piece?
This a question we are all familiar with and anyone who has seen it anywhere/solved it knows that the expected length of the medium piece is $\frac{5}{18}$, we reach this conclusion by using $E(L+M+S)=1$, and we know we can calculate E(L) and E(S), so we just use $E(M)=1-<!--\; -\; -->E(L)-<!--\; -\; -->E(S)$ to get our answer, is there any way we can solely calculate the E(M) without calculating the other two values?
L: Length of the longest part
S: Length of the smallest part
M: Length of the medium part

What is the distance between the origin of a cartesian coordinate system and the point (-3,7)?

Distance between (-6,3,1) and (0,4,-2)?

Is there any relationship similar to the following. Let $X$ be the maximum of functions $f_1(x)+f_2(x)$. Let $X_1$ be a maximum of $f_1(x)$ and let $X_2$ be a maximum of $f_2(x)$. Is there any relationship between $X$ and $X_1$ and $X_2$?

For example can we say under what condition will $X$ be in-between $X_1$ and $X_2$, $X_1\le <!--\; \le \; -->X\le <!--\; \le \; -->X_2$.

Any reference would be greatly appreciated.

Probability of geometric Brownian motion
I want to solve the following question: Let X(t) be the price of JetCo stock at time t years from the present. Assume that X(t) is a geometric Brownian motion with zero drift and volatility $\mathrm{\sigma <!--\; \sigma \; -->}=0.4/yr$. If the current price of JetCo stock is 8.00 USD, what is the probability that the price will be at least 8.40 USD six months from now.
Here is my attempt: Since X(t) is a geometric Brownian motion, log(X(t)) is a regular Brownian motion with zero drift and $\mathrm{\sigma <!--\; \sigma \; -->}=0.4/yr$. We want to find the probability that $\log <!--\; \; -->(X(1/2))\ge <!--\; \ge \; -->\log <!--\; \; -->(8.40)$ given that $\log <!--\; \; -->(X(0))=\log <!--\; \; -->(8.00)$.
What can I do from here?

Polygons with two different angles
I have the following questions:
Let us consider polygons with only two different interior angles $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}=\mathrm{\pi <!--\; \pi \; -->}-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}$.
For which frequencies of these angles is it possible to form a closed polygon?

In this image, $MN\parallel <!--\; \parallel \; -->AT$, $CD=DA$ and $\mathrm{\angle <!--\; \angle \; -->}PDT=90º$ then arc $NP$=?

Volume of Generalized Tetrahedron in $R^n$.
I'm having difficulty finding the volume of a tetrahedron in ${\mathbb{R}}^n$.
Find the volume of a generalized tetrahedron in ${\mathbb{R}}^n$ bounded by the coordinate hyperplanes and the hyperplane $x_1+x_2+...+x_n=1$
In two dimensions, we have ${\int <!--\; \int \; -->}_0^{1}1-<!--\; -\; -->x_{1}d{x}_1$. In three dimensions, I got something like ${\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x_1}1-<!--\; -\; -->x_{2}d{x}_{2}d{x}_1$.
I am off to a good start?

Distance between (-2,2,6) and (-5,-1,-1)?

Finding the volume of the tetrahedron using triple integrals
D is the tetrahedron bounded by the coordinate planes and the plane $3x+3y+z=3$, then express the volume of D as a triple integral.
My Try:
The z-limits are $0\le <!--\; \le \; -->z\le <!--\; \le \; -->3-<!--\; -\; -->3x-<!--\; -\; -->3y$
If y is the limit on the xy- plane then $z=0$ $0\le <!--\; \le \; -->y\le <!--\; \le \; -->1-<!--\; -\; -->x$.
Similarly, $0\le <!--\; \le \; -->x\le <!--\; \le \; -->3$.
Finally, the integral will be ${\int <!--\; \int \; -->}_0^3{\int <!--\; \int \; -->}_0^{1-<!--\; -\; -->x}{\int <!--\; \int \; -->}_0^{3-<!--\; -\; -->3x-<!--\; -\; -->3y}dzdydx$

Finding the volume by Shell method: $y=x^2,y=2-<!--\; -\; -->x^2$, $y=x^2,y=2-<!--\; -\; -->x^2$ about the line $x=1$.
what I get from this after graphing is:
$2\mathrm{\pi <!--\; \pi \; -->}\int <!--\; \int \; -->(1-<!--\; -\; -->x)(2-<!--\; -\; -->2x^2)dx$
which becomes: $2\mathrm{\pi <!--\; \pi \; -->}\int <!--\; \int \; -->(2-<!--\; -\; -->2x^2-<!--\; -\; -->2x+2x^3)dx$
integrating that I get:
$2\mathrm{\pi <!--\; \pi \; -->}[2x-<!--\; -\; -->\frac{2}{3}{x}^3-<!--\; -\; -->x^2+\frac{1}{2}{x}^4]$ from 0 to 1
My answer is $\frac{5}{3}\mathrm{\pi <!--\; \pi \; -->}$ but my book says the answer is: $\frac{16}{3}\mathrm{\pi <!--\; \pi \; -->}$
Could someone tell me where I went wrong? Was it the upper lower bounds?

Two people play a game. They play a series of points, each producing a winner and a loser, until one player has won at least 4 points and has won at least 2 more points than the other. Anne wins each point with probability p. What is the probability that she wins the game on the kth point played for $k=4,5,6,...$