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Let \(F_1=(10,2)\) and  \(F_2 = (-16,2)\). Then the set of points \(P\) such that \(|PF_1 - PF_2| = 24\) form a hyperbola.
The equation of this hyperbola can be written as \(\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1\).
Find \(h + k + a + b\).

\(\begin{array}{|rcll|} \hline \mathbf{|PF_1 - PF_2|} &=& \mathbf{2a} \\ |PF_1 - PF_2| &=& 24 \\ 2a&=& 24 \\ \mathbf{a} &=& \mathbf{12} \\ \hline \end{array}\)

\(\begin{array}{|lrc|} \hline \mathbf{\text{Focus}_1:}& \mathbf{F_1(h+ae,k)} \\ & F_1(10,2): & h+ae = 10 \quad (1) \\ && \mathbf{ k =2} \\ \hline \mathbf{\text{Focus}_2:}& \mathbf{F_2(h-ae,k)} \\ & F_2(-16,2): & h-ae = -16\quad (2) \\ && \mathbf{ k =2} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1)+(2): & h+ae + h-ae &=& 10-16 \\ & 2h &=& -6 \\ & \mathbf{ h }&=& \mathbf{-3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1) & h+ae &=& 10 \quad |\quad h=-3 \\ & -3+ae &=& 10 \\ & \mathbf{ ae }&=& \mathbf{13} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a^2 + b^2 &=& (ae)^2 \\ 12^2+b^2 &=& 13^2 \\ b^2 &=& 13^2-12^2 \\ b^2 &=& 169-144 \\ b^2 &=& 25 \\ \mathbf{ b }&=& \mathbf{5} \\ \hline \end{array}\)

\(\begin{array}{rcll} && \mathbf{h + k + a + b} \\ &=& -3 + 2 +12+5 \\ &=& \mathbf{16} \\ \end{array}\)

The asymptotes of a hyperbola are \(y = 2x - 3\) and \(y = 17 - 2x\).
Also, the hyperbola passes through the point \((4,7)\).
Find the distance between the foci of the hyperbola.

Formula hyperbola:

\(\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ asymptote}} & 2x - 3 &=& y \\ & 2x - 3-y &=& 0 \\\\ \hline \mathbf{2. \text{ asymptote}} & 17 - 2x &=& y \\ & 17 - 2x-y &=& 0 \\\\ \hline \text{hyperbola: } & (2x - 3-y)(17 - 2x-y) + c &=& 0 \\\\ P(x=4,y=7) & (8-3-7)(17-8-7)+c &=& 0 \\ & (-2)(2)+c &=& 0 \\ & c &=& 4 \\ \mathbf{\text{hyperbola: }} &\mathbf{ (2x - 3-y)(17 - 2x-y) + 4} &\mathbf{=}& \mathbf{0} \\\\ & (2x - 3-y)(17 - 2x-y) + 4 &=& 0 \\ & 34x-4x^2-2xy-51+6x+3y-17y+2xy+y^2+4 &=& 0 \\ & -4x^2+40x+y^2-14y &=& 47 \\ &\ldots \\ & \mathbf{ \dfrac{(x-5)^2}{1} - \dfrac{(y-7)^2}{4}} &=& \mathbf{ 1 } \\ & \boxed{\mathbf{\text{Formula hyperbola:} } \dfrac{ (x-h)^2 }{a^2}- \dfrac{ (x-k)^2 }{b^2} = 1 } \\\\ & a^2 = 1 \\ & \mathbf{a= 1} \\\\ & b^2 = 4 \\ & \mathbf{b= 2} \\ \hline \end{array}\)

\(\begin{array}{|l|rcll|} \hline \text{Focus}_1: & F_1(h+ae,k) \\ \text{Focus}_2: & F_2(h-ae,k) \\ \hline \text{Foci distance}: & && h+ae-(h-ae) \\ & &=&h+ae-h+ae \\ & &=& 2ae \quad | \quad e=\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2a\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=&\mathbf{ 2 \sqrt{a^2+b^2}} \quad | \quad a^2=1, \quad b^2= 4 \\ & &=& 2 \sqrt{1+4} \\ & &\mathbf{=}& \mathbf{2\sqrt{5} } \\ \hline \end{array} \)

The distance between the foci of the hyperbola is \(\mathbf{2\sqrt{5} }\)

For a certain hyperbola \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]where a > b, the angle between the asymptotes is \( 60^\circ\). Find \(\frac{a}{b}\).

Hello Guest!

\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

\(y=b\cdot \sqrt{\frac{x^2}{a^2}-1}\)

For \(x→(-\ ∞)\) the following applies:

\(y\ '=\pm\ tan(30°)\)

\(y\ '=b\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{\frac{x^2-a^2}{a^2}}}\cdot\frac{2x}{a^2}\\ y\ '=b\cdot \frac{1}{2}\cdot \sqrt{\frac{a^2}{x^2-a^2}}\cdot\frac{2x}{a^2}\\ y\ '=\frac{bx}{a^2}\cdot \sqrt{\frac{a^2}{x^2-a^2}}\\ y\ '=\frac{b}{a}\cdot \sqrt{\frac{x^2}{x^2-a^2}}\\\)

\(For\ x→(-\ ∞)\ the\ following\ applies:\)

\(y\ '=\frac{b}{a}\cdot \sqrt{\frac{x^2}{x^2-a^2}}=tan(30°)\)

\(For\ x→(-\ ∞):\\ \sqrt{\frac{x^2}{x^2-a^2}}=1\)

\(\frac{b}{a}=tan(30°)\)

\(\frac{b}{a}= \frac{1}{3}\sqrt{3}\\ \frac{a}{b}=\pm \sqrt{3}\)

\(\color{BrickRed}a>b\\ \color{blue} \frac{a}{b}=\sqrt{3}\)

  !

Don't you do any of your own homework?

Let P be the intersection point of the line through points D = (1, 1, 2) and E = (2, 3, 4) 

with the plane through A = (0,1,1), B = (1,1,0) and C = (1,0,3).

What is P?

\(\begin{array}{|rcl|rclrcl|} \hline \text{plane} &&& \text{line} \\ \hline && A = \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix} \qquad B = \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} \qquad C = \begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix} \qquad & && D = \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} \qquad E = \begin{pmatrix} 2\\ 3\\ 4 \end{pmatrix} \\ \vec{x} &=& \vec{B}+s(\vec{A}-\vec{B})+t(\vec{C}-\vec{B}) & \vec{x} &=& \vec{D}+r(\vec{E}-\vec{D}) \\\\ \vec{x} &=&\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} +s\left(\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}\right) +t\left(\begin{pmatrix} 1\\ 0\\ 3 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}\right) & \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix} +r\left(\begin{pmatrix} 2\\ 3\\ 4 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}\right) \\\\ \vec{x} &=&\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} & \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \vec{x} =\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}+s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} &=&\begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} -r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}-\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} \\\\ \mathbf{ s\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}+t\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix} -r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} } &=& \mathbf{ \begin{pmatrix} 0\\ 0\\ 2 \end{pmatrix} } \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & -s-r &=& 0 \\ & \mathbf{r} &=& \mathbf{ -s} \\ \hline (2) & -t-2r &=& 0 \\ & t &=& -2r \\ & t &=& -2(-s) \\ & \mathbf{t} &=& \mathbf{2s} \\ \hline (3) & s+3t-2r &=& 2 \\ & s +3(2s)-2(-s) &=& 2 \\ & s+6s+2s &=& 2 \\ & 9s &=& 2 \\ & \mathbf{s} &=& \mathbf{\dfrac{2}{9}} \\\\ & t &=& 2s \\ & t &=& 2\left(\dfrac{2}{9}\right) \\ &\mathbf{t} &=& \mathbf{\dfrac{4}{9}} \\\\ & r &=& -s \\ & r &=& -\left(\dfrac{2}{9}\right)\\ &\mathbf{r} &=& \mathbf{- \dfrac{2}{9} } \\ \hline \end{array}\)

\(\mathbf{\vec{P}=\ ?}\)

\(\begin{array}{|rcll|} \hline \vec{x} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}+r\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \vec{P} &=& \begin{pmatrix} 1\\ 1\\ 2 \end{pmatrix}- \dfrac{2}{9}\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix} \\\\ \vec{P} &=& \begin{pmatrix} 1-\dfrac{2}{9}\\ 1-\dfrac{4}{9}\\ 2-\dfrac{4}{9} \end{pmatrix} \\\\ \mathbf{\vec{P}} &=& \mathbf{\begin{pmatrix} \dfrac{7}{9}\\ \dfrac{5}{9}\\ \dfrac{14}{9} \end{pmatrix} } \\ \hline \end{array} \)

see correction below

Thanks for that addition Ginger,

I remember looking into the night sky to see if I could spot Apollo 11.

I have lived in a wondrous era !

I wonder what the next century has in store for the children of today! 

Can technological advances continue at the same acceleration?  My mind bogges.

Next year we are getting a air taxi service in Melbourne.  The craft will land on many key buiding tops. 

soon it will be like the Jetson's!

The sum of the digits in each expression is 15. 

7 + 3 + 5 = 15

8 + 4 + 3 = 15

9 + 5 + 1 = 15

If this is the right way to solve the problem, then 6 + 5 + 4 = 15 which gives us the answer is 4. 

.

6+19+30 = 55 appears to be the minimum value that satisifies the conditions.

6+19 = 25

6+30 = 36

19=30=49

Thanks Rom :)

It is good to see that our answers are getting looked at   

Thanks Rom and guest.

thanks 

Thanks!!!!

\(\text{There are a couple ways to do this. We'll do it formally first, by setting the gradient to 0}\\ f=\dfrac{a}{2b}+\dfrac{b}{4c}+\dfrac{c}{8a}\\ \nabla f = \left(\dfrac{1}{2b}-\dfrac{c}{8a^2},~\dfrac{1}{4c}-\dfrac{a}{2b^2},~\dfrac{1}{8a}-\dfrac{b}{4c^2}\right)\\ \nabla f=0 \Rightarrow\\ \dfrac{1}{2b}=\dfrac{c}{8a^2}\\ \dfrac{1}{4c}=\dfrac{a}{2b^2}\\ \dfrac{1}{8a}=\dfrac{b}{4c^2}\)

\(\text{The only solution with no negative values is $b=2a,~c=2a$}\\ \text{This results in $f = \dfrac 3 4$ but more importantly it results in $~\\$ each of the terms contributing equally to the sum}\\ f = \dfrac{a}{4a}+\dfrac{2a}{8a} + \dfrac{2a}{8a} = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4\)

\(\text{You can use that symmetry principle to come up with the answer in a simpler way}\\ \text{Just find the relationship between the variables that makes each term of what you're trying to $~\\$ find the extrema of contribute the same amount}\\ \text{Here you'd say that $\dfrac{a}{2b} = \dfrac{b}{4c} = \dfrac{c}{8a}$}\\ \text{and solve for $b$ and $c$ in terms of $a$ as we did above}\)

In all these symmetric problems like this, with non-negative elements, either all the elements are equal or all but one element is zero.

We just have to figure out which corresponds to the min and which to the max.

If a,b,c=1/3, each term in the sum is 1/6, so the sum is 1/2

if a=1, b=c=0, oops can't do that, we need at least 2 non-zero terms.

Suppose a=b=1/2, c=0

1/4 + 0+0 = 1/4 < 1/2

So the maximum occurs when all the variables are equal at 1/3 and is equal to 1/2.

You can show this formally using Lagrange Multipliers if you like.

I'm not sure what method they intend for you to use.

I suspect they want you to unmangle the expression so that the Cauchy-Schwarz inequality can be applied.

^{... UNfortunately....we get a lot of folks posting stuf which is UNclear!   Parentheses and brackets are NECESSARY to get correct answers to postings (and not make an engineering mistake that kills people !)}

^------

I agree that the postings from many students are unclear. It’s obvious these students lack the prerequisites for the posted questions; including an understanding of basic mathematical hierarchical conventions. Without these prerequisites, they will not understand where to place parenthetical operators, or they believe them to be unnecessary even when included in the original question.

These deficits point to an apparent flaw in lower-level education of students for these hierarchical conventions. Based on my observations, the debate on these conventions are usually by second-year college/university students and reaches a crescendo every other year. It may be because the crescendo carries over to first year students, that when they become second-year students it’s mostly a moot point and last year’s news. 

As for engineers, especially those who have the credentials and experience to create or certify anything that might cause death to a population, or the destruction of expensive science experiments, it’s reasonable to assume they are well versed on such hierarchies. In addition, all mathematics and algorithms are peer reviewed and subjected to tests to catch transient math and logic errors.  This process works well, as indicated by the rare failure. Most failures, when they occur, are usually traceable to a sequence of errors where no single event by itself would have caused the failure.

In modern computing, redundant uses of parenthetical operators are a trivial matter, but in the early days of computerized control, mnemonic and variable storage space was always at a premium and maxed-out, so superfluous parenthetical operators were not an option. 

The article below gives insight to the computer related testing and contingency plans in use for the Apollo 11 moon landing.

\(w_1,w_2 \in \mathbb{C} \Rightarrow w_1w_2 = |w_1||w_2|e^{\arg(w_1)+\arg(w_2)}\\ \text{So in determining what quadrant the result is in we add the individual angles}\)

\(\arg(z_1) \approx \dfrac \pi 6\\ \arg(z_2) \approx \dfrac \pi 4\\ \arg(z_3) \approx \dfrac{7\pi}{12}\\ \arg(z_4) \approx \dfrac{7\pi}{6}\)

\(\text{Now adding these according to the products you're given}\\ \begin{array}{ccccc} \text{Product}&\text{angle 1}&\text{angle 2}&\text{sum}&\text{quadrant}\\ z_1z_2 &\dfrac\pi 6&\dfrac \pi 4 &\dfrac{5\pi}{12} &1\\ z_1z_3 &\dfrac\pi 6&\dfrac{7\pi}{12} &\dfrac{9\pi}{12} &2\\ z_1z_4 &\dfrac\pi 6&\dfrac{7 \pi}{6} &\dfrac{4\pi}{3} &3\\ z_2z_3 &\dfrac\pi 4&\dfrac {7\pi}{12} &\dfrac{5\pi}{6} &2\\ z_2z_4 &\dfrac\pi 4&\dfrac {7\pi}{6} &\dfrac{17\pi}{12} &3\\ z_3z_4 &\dfrac{7\pi}{12}&\dfrac {7\pi}{6} &\dfrac{7\pi}{4} &4\\ \end{array}\)

I checked the answer via Mathematica as usual doing

Select[Range[100000, 999999], 
  Plus @@ IntegerDigits[#] == 23 &] // Length

This selects all integers in the range 100000-999999

whose sum of digits equals 23 and then counts the length of that list.

It returns 41712

As I said before I didn't consider this a single problem.

When we apply all 3 conditions yes the answer is 352

I'm going to have to give some thought (or not) about how to derive that number w/o software.

Let A=(10,-10) and O=(0,0).

Determine the sum of all x and y-coordinates of all points Q on the line y=-x+6

such that angle OQA = 90.

\(\begin{array}{|rcll|} \hline -\dbinom{x_Q}{y_Q}\dbinom{x_A-x_Q}{y_A-y_Q} &=& 0 \\\\ \dbinom{x_Q}{y_Q}\dbinom{x_A-x_Q}{y_A-y_Q} &=& 0 \quad | \quad y_Q = 6-x_Q \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{x_A-x_Q}{y_A-(6-x_Q)} &=& 0 \quad | \quad x_A = 10,\ y_A=-10 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-10-(6-x_Q)} &=& 0 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-10-6+x_Q} &=& 0 \\\\ \dbinom{x_Q}{6-x_Q}\dbinom{10-x_Q}{-16+x_Q} &=& 0 \\\\ x_Q(10-x_Q) + (6-x_Q)(-16+x_Q) &=& 0 \\ 10x_Q-x_Q^2 -96+6x_Q+16x_Q-x_Q^2 &=& 0 \\ -2x_Q^2 + 32x_Q-96 &=& 0 \quad | \quad : (-2) \\ \mathbf{x_Q^2 - 16x_Q + 48} &=& \mathbf{0} \\ x_Q &=& \dfrac{16\pm \sqrt{16^2-4\cdot 48} }{2} \\ x_Q &=& \dfrac{16\pm \sqrt{64} }{2} \\ x_Q &=& \dfrac{16\pm 8 }{2} \\ x_Q &=& 8 \pm 4 \\\\ x_Q &=& 8+4 \\ \mathbf{x_Q} &=& \mathbf{12} \\ y_Q &=& 6-x_Q \\ y_Q &=& 6-12 \\ \mathbf{y_Q} &=& \mathbf{-6} \\\\ x_Q &=& 8-4 \\ \mathbf{x_Q} &=& \mathbf{4} \\ y_Q &=& 6-x_Q \\ y_Q &=& 6-4 \\ \mathbf{y_Q} &=& \mathbf{2} \\ \hline \end{array}\)

The sum of all x and y-coordinates of all points Q
\(12-6+4+2 = \mathbf{12}\)

duh, right you are.

it's just \([4, \infty)\)

lcm(33, 45)/gcd(33, 45)=495/3=165

Thanks for the help!!!

Thanks!

Thanks!!!