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 #1
avatar+26364 
+2

I don't get eccentricity and its relation to hyperbolas...in a question where I have to form an equation for a hyperbola with a center at (-3,1), one focus at (2,1), and eccentricity is 5/4, how would I solve this? 

 

\(center : ~(h=-3,~ k=1)\)

\(focus: ~( x_f = 2,~ y_f=1)\)

\(eccentricity: ~e = \frac{5}{4}\)

 

Because \(k = y_f =1\) the transverse axis is horizontal.

 

1.  c = ?

\(\begin{array}{|rcll|} \hline focus: ( h+c,~k ) &=& (2,1)\\ h+c &=& 2 \\ -3+c &=& 2 \\ c&=& 2+3\\ \mathbf{c} &\mathbf{=}& \mathbf{5} \\ \hline \end{array}\)

 

(2.)

\(\begin{array}{|rcll|} \hline focus_2: ( h-c,~k ) &=& (x_f,y_f=k)\\ h-c &=& x_f \\ -3-5 &=& x_f \\ x_f &=& -8\\ y_f &=& 1 \\ \mathbf{focus_2} &\mathbf{=}& \mathbf{(-8,1)} \\ \hline \end{array} \)

 

3. a = ?

\(\begin{array}{|rcll|} \hline \mathbf{e} &\mathbf{=}& \mathbf{\frac{c}{a}} \\ \frac{5}{4} &=& \frac{5}{a} \\ \frac{4}{5} &=& \frac{a}{5} \\ a &=& \frac{4}{5}\cdot 5 \\ \mathbf{a} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

 

4. b = ?

\(\begin{array}{|rcll|} \hline \mathbf{c^2} &\mathbf{=}& \mathbf{a^2+b^2} \\ b^2 &=&c^2-a^2 \\ b^2 &=&5^2-4^2 \\ b^2 &=&25-16 \\ b^2 &=& 9 \\ \mathbf{b} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}\)

 

5.  Equation:

\(\begin{array}{|rcll|} \hline \mathbf{\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}} &\mathbf{=}& \mathbf{1} \\ \frac{(x-(-3))^2}{4^2}-\frac{(y-1)^2}{3^2} &=& 1 \\ \mathbf{\frac{(x+3)^2}{4^2}-\frac{(y-1)^2}{3^2}} &\mathbf{=}& \mathbf{1} \\ \hline \end{array} \)

 

6. The graph:

 

 

laugh

22.01.2018
 #6
avatar+26364 
0

Compute the sum    2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5)+...

 

see link: http://web2.0calc.com/questions/algebra_47009

 

In General:

\(\begin{array}{lcll} s = \dfrac{1}{1 \cdot 2 } + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4 } + \dfrac{1}{4 \cdot 5 } + \cdots \ + \dfrac{1}{n \cdot (n+1)} + \cdots =\ \frac{1}{1!0!}\cdot \frac{1}{1} = 1 \\\\ s =\mathbf{ \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \frac{1}{2!0!}\cdot \frac{1}{2} } = \frac{1}{4} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3)} + \cdots =\ \frac{1}{3!0!}\cdot \frac{1}{3} = \frac{1}{18} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3 \cdot (n+4) } + \cdots =\ \frac{1}{4!0!}\cdot \frac{1}{4} = \frac{1}{96} \\\\ \ldots \\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdots } + \dfrac{1}{2 \cdot 3 \cdot 4 \cdots } + \dfrac{1}{3 \cdot 4 \cdot 5 \cdots} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)\cdot~\cdots ~\cdot (n+m) } + \cdots =\ \frac{1}{m!0!}\cdot \frac{1}{m} = \frac{1}{m\cdot m!} = \frac{1}{(m+1)!-m!} \\\\ \end{array} \\ \)

 

 

 

 

laugh

19.01.2018
 #3
avatar+26364 
+1

A circle of radius 5 with its center at $(0,0)$ is drawn on a Cartesian coordinate system.

How many lattice points (points with integer coordinates) lie within or on this circle?

 

A Calculation of the Number of Lattice Points within or on the circle:

 

Let \( \lfloor x \rfloor \) be the largest integer equal to or less than x.

 

Example:
\(\lfloor 3.53553390593 \rfloor = 3\)
\(\lfloor -3.53553390593 \rfloor = -4\)

 

 

 

Noted by Gauss:

Let r  radius of the circle = 5

Let \(x = r^2\)

 

\(\begin{array}{|rcll|} \hline A_2(x) &=& 1 + 4\lfloor \sqrt{x} \rfloor + 4 \lfloor \sqrt{\frac{x}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{x}{2}} \rfloor + 1 }^{\lfloor \sqrt{x} \rfloor} \lfloor \sqrt{x-y_1^2} \rfloor \qquad & | \quad x = r^2 = 5^2 \\\\ &=& 1 + 4\lfloor \sqrt{5^2} \rfloor + 4 \lfloor \sqrt{\frac{5^2}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{5^2}{2}} \rfloor + 1 }^{\lfloor \sqrt{5^2} \rfloor} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 3 + 1 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 4 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( \lfloor \sqrt{5^2-4^2} \rfloor +\lfloor \sqrt{5^2-5^2} \rfloor \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( 3 + 0 \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 24 \\\\ &=& 1 + 20 + 36 + 24 \\ &\mathbf{=} & \mathbf{81} \\ \hline \end{array}\)

 

81 lattice points (points with integer coordinates) lie within or on this circle with radius 5.

 

Example:
\(r = 0 \ldots 20\)

 

Number of lattice points in circle:

\(\begin{array}{|r|r|r|} \hline r & \text{lattice points in circle} & \text{lattice points in sphere } \\ \hline 0 & 1 & 1 \\ 1 & 5 & 7 \\ 2 & 13 & 33 \\ 3 & 29 & 123 \\ 4 & 49 & 257 \\ {\color{red}5} & {\color{red}81} & 515 \\ 6 & 113 & 925 \\ 7 & 149 & 1419 \\ 8 & 197 & 2109 \\ 9 & 253 & 3071 \\ 10 & 317 & 4169 \\ 11 & 377 & 5575 \\ 12 & 441 & 7153 \\ 13 & 529 & 9171 \\ 14 & 613 & 11513 \\ 15 & 709 & 14147 \\ 16 & 797 & 17077 \\ 17 & 901 & 20479 \\ 18 & 1009 & 24405 \\ 19 & 1129 & 28671 \\ 20 & 1257 & 33401 \\ \hline \end{array} \)

 

laugh

15.01.2018