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 #16
avatar+26400 
+25

Hallo Bertie,

\(60y^2 + 65x^2 - 128xy - 448x +448y + 768+{\color{red}60\cdot c}= 0 \qquad c > 0 \)   hyperbola between lines

 

Here is the answer:

 

I.

\(\small{ \begin{array}{lrcll} \mathbf{\text{1. line}} & 13x-10y-48=0 \\ &y &=& m_1x+b_1\\ &y &=&\frac{13}{10}x -\frac{48}{10} \\ & m_1 &=& \frac{13}{10}\\ & b_1 &=& -\frac{48}{10}\\ \text{or } & m_1x+b_1-y &=& 0 \qquad (\frac{13}{10}x -\frac{48}{10}-y=0) \\ \\ \mathbf{\text{2. line}} & 5x-6y-16=0 \\ &y &=& m_2x+b_2\\ &y &=&\frac{5}{6}x -\frac{16}{6} \\ & m_2 &=& \frac{5}{6}\\ & b_2 &=& -\frac{16}{6}\\ \text{or } & m_2x+b_2-y &=& 0 \qquad (\frac{5}{6}x -\frac{16}{6} -y=0) \\\\ \mathbf{\text{Intersection 1. line}} \\ \mathbf{\text{and 2.line}}\\ & x_c &=& \frac{b_2-b_1}{m_1-m_2} = \frac{32}{7} \\ & y_c &=& \frac{m_1b_2-b_1m_2}{m_1-m_2} = \frac{8}{7} \\ \end{array} }\)

 

II. The Hyperbola is the product of the two lines + c : \(\boxed{~ (m_1x+b_1-y)(m_2x+b_2-y) + c = 0 ~}\)   c > 0

 

\(\small{ \begin{array}{lrcll} \text{or } &\boxed{~ y^2+m_1m_2x^2+\left(m_1b_2+m_2b_1\right)x-\left(m_1+m_2\right)xy-\left(b_1+b_2\right)y+b_1b_2+c=0~} \\ \text{or } & y^2+ \frac{13}{12}x^2-\frac{112}{15}x-\frac{32}{15}xy+\frac{112}{15}y+\frac{64}{5}+c=0 \qquad \cdot 60\\ \text{or } &\boxed{~ 60y^2+ 65x^2-448x-128xy+448y+768+60\cdot c=0 ~} \end{array} }\)

 

 

III. The Hyperbola is:  \(\small{ \begin{array}{rll} Y^2 + aX^2 + bXY + c &=& 0. \qquad | \quad X = x - x_c,~ Y = y - y_c & [see\ \#6]\\ \end{array} } \)

\(\small{ \begin{array}{lclcrl} a &=& m_1m_2 &=&\frac{13}{13} & [see\ \#11]\\ b &=& -(m_1+m_2) &=& -\frac{32}{15} & [see\ \#11]\\ x_c &=& \frac{b_2-b_1}{m_1-m_2} &=& \frac{32}{7} &\\ y_c &=& \frac{m_1b_2-b_1m_2}{m_1-m_2} &=& \frac{8}{7} & \\ \end{array} }\)

 

let us put all things together:

\(\small{ \begin{array}{rcll} (y - \frac{m_1b_2-b_1m_2}{m_1-m_2})^2 + m_1\cdot m_2\cdot (x - \frac{b_2-b_1}{m_1-m_2})^2 - (m_1+m_2)\cdot (x - \frac{b_2-b_1}{m_1-m_2})\cdot (y - \frac{m_1b_2-b_1m_2}{m_1-m_2}) + c &=& 0\\ \cdots \\ \end{array} }\\ \)

\(\small{ \begin{array}{lcll} &=& y^2\\ &+&m_1m_2x^2\\ &+&\left[ \underbrace{ \frac{(m_1+m_2)(m_1b_2-b_1m_2)-2m_1m_2(b_2-b_1)}{m_1-m_2}}_{1.} \right] x\\ &-&(m_1+m_2)xy\\ &+&\left[ \underbrace{ \frac{(m_1+m_2)(b_2-b_1)-2(m_1b_2-b_1m_2)}{m_1-m_2}}_{2.}\right] y\\ &+&\underbrace{ \frac{(m_1b_2-b_1m_2)^2+m_1m_2(b_2-b_1)^2-(m_1+m_2)(b_2-b_1)(m_1b_2-b_1m_2)}{(m_1-m_2)^2}}_{3.}\\ &+& c = 0 \end{array} }\)

 

calculate 1. :

\(\small{ \begin{array}{lcll} \frac{(m_1+m_2)(m_1b_2-b_1m_2)-2m_1m_2(b_2-b_1)}{m_1-m_2} \\ = \frac{(m_1+m_2)m_1b_2 -(m_1+m_2)b_1m_-2m_1m_2b_2+2m_1m_2b_1}{m_1-m_2} \\ \cdots \\ = \frac{-m_1m_2b_2+m_1m_2b_1+m_1^2b_2-m_2^2b_1}{m_1-m_2}\\ = \frac{b_1m_2(m_1-m_2)+b_2m_1(m_1-m_2)}{m_1-m_2}\\ \mathbf{= b_1m_2+b_2m_1}\\ \end{array} }\)

 

calculate 2. :

\(\small{ \begin{array}{lcll} \frac{(m_1+m_2)(b_2-b_1)-2(m_1b_2-b_1m_2)}{m_1-m_2}\\ = \frac{ m_1b_2-m_1b_1+m_2b_2-m_2b_1-2m_1b_2+2b_1m_2}{m_1-m_2}\\ = \frac{ -m_1b_2+b_1m_2-m_1b_1+m_2b_2}{m_1-m_2}\\ = \frac{-b_2(m_1-m_2)-b_1(m_1-m_2)}{m_1-m_2}\\ = -b_2-b_1\\ \mathbf{= -(b_1+b_2) }\\ \end{array} }\)

 

calculate 3. :

\(\small{ \begin{array}{lcll} \frac{(m_1b_2-b_1m_2)^2+m_1m_2(b_2-b_1)^2-(m_1+m_2)(b_2-b_1)(m_1b_2-b_1m_2)}{(m_1-m_2)^2} \\ \cdots \\ = \frac{-2m_1m_2b_1b_2+m_1^2b_1b_2+m_2^2b_1b_2}{(m_1-m_2)^2} \\ = \frac{b_1b_2(m_1^2-2m_1m_2+m_2^2)}{(m_1-m_2)^2} \\ = \frac{b_1b_2(m_1-m_2)^2)}{(m_1-m_2)^2} \\ \mathbf{= b_1b_2} \\ \end{array} }\)

 

So we have the same:

\(\small{ \begin{array}{lrcll} \boxed{~ y^2+m_1m_2x^2+\left(m_1b_2+m_2b_1\right)x-\left(m_1+m_2\right)xy-\left(b_1+b_2\right)y+b_1b_2+c=0~} \\ \end{array} }\)

 

laughlaughlaugh

Jan 22, 2016
 #2
avatar
+5

Simplify the following:
((p^2 q)^3 ((2 p^3)/q^4)^2)/(p^2 q^3 ((4 q^5)/p^2)^3)

Multiply each exponent in p^2 q by 3:
(p^(3×2) q^3 ((2 p^3)/q^4)^2)/(p^2 q^3 ((4 q^5)/p^2)^3)

3×2  =  6:
(p^6 q^3 ((2 p^3)/q^4)^2)/(p^2 q^3 ((4 q^5)/p^2)^3)

Multiply each exponent in (2 p^3)/q^4 by 2:
(p^6 q^3×(2^2 p^(2×3))/((q^4)^2))/(p^2 q^3 ((4 q^5)/p^2)^3)

2×3  =  6:
(2^2 p^6 q^3 p^6)/((q^4)^2 p^2 q^3 ((4 q^5)/p^2)^3)

2^2 = 4:
(4 p^6 q^3 p^6)/((q^4)^2 p^2 q^3 ((4 q^5)/p^2)^3)

Multiply exponents. (q^4)^2 = q^(4×2):
(4 p^6 q^3 p^6)/(q^(4×2) p^2 q^3 ((4 q^5)/p^2)^3)

4×2  =  8:
(4 p^6 q^3 p^6)/(q^8 p^2 q^3 ((4 q^5)/p^2)^3)

Multiply each exponent in (4 q^5)/p^2 by 3:
(4 p^6 q^3 p^6)/(q^8 p^2 q^3×(4^3 q^(3×5))/((p^2)^3))

3×5  =  15:
(4 p^6 q^3 p^6)/(q^8×(4^3 p^2 q^3 q^15)/(p^2)^3)

Multiply exponents. (p^2)^3 = p^(2×3):
(4 p^6 q^3 p^6)/(q^8×(4^3 p^2 q^3 q^15)/p^(2×3))

2×3  =  6:
(4 p^6 q^3 p^6)/(q^8×(4^3 p^2 q^3 q^15)/p^6)

4^3 = 4×4^2:
(4 p^6 q^3 p^6)/(q^8×(4×4^2 p^2 q^3 q^15)/p^6)

4^2 = 16:
(4 p^6 q^3 p^6)/(q^8×(4×16 p^2 q^3 q^15)/p^6)

4×16  =  64:
(4 p^6 q^3 p^6)/(q^8×(64 p^2 q^3 q^15)/p^6)

(p^6 q^3×4 p^6)/(q^8 (p^2 q^3×64 q^15)/p^6) = q^3/q^3×((p^6×4 p^6)/q^8)/((p^2×64 q^15)/p^6) = ((p^6×4 p^6)/q^8)/((p^2×64 q^15)/p^6):
((4 p^6 p^6)/q^8)/((64 p^2 q^15)/p^6)

Combine powers. (p^6×4 p^6)/(q^8 (p^2×64 q^15)/p^6) = (p^(6+6-2) 4 p^6)/(q^8×64 q^15):
(4 p^6+6-2 p^6)/(64 q^8 q^15)

6+6-2 = 10:
(4 p^10 p^6)/(64 q^8 q^15)


4 | 1 | 6
| 6 | 4
- | 4 |
| 2 | 4
- | 2 | 4
|  | 0:
(p^10 p^6)/(16 q^8 q^15)

Combine powers. (p^10 p^6)/(16 q^8 q^15) = (p^(10+6) q^(-8-15))/16:
(p^10+6 q^-8-15)/16

10+6 = 16:
(p^16 q^(-8-15))/16

-8-15 = -23:
Answer: | (p^16 q^-23)/16

Jan 22, 2016
 #3
avatar+4084 
+5

Ur welcome!

Jan 22, 2016
 #2
avatar+600 
+10

Lol. Thx.

Jan 22, 2016
 #1
avatar+4084 
0

Google....🙃

Jan 22, 2016
 #2
avatar
0

ok thanks

Jan 22, 2016

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