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Write the equation of hyperbole which is tangent with the straights 5x-6y-16=0 and 13x-10y-48=0.

Guest Jan 18, 2016

Best Answer 

 #16
avatar+18712 
+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

heureka  Jan 22, 2016
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17+0 Answers

 #1
avatar+2493 
+5

i think:

 

let s found what does your y s equal to:

5x-6y-16=0                               13x-10y-48=0                   

6y=5x-16                                  10y=13x-48

y=5x/6-16/6                               y=13x/10-48/10

 

lets found the point intersection of theese points:

 

5x/6-16/6=13x/10-48/10

x = 32/7

put x into one of the equations to find y 

 

y=(5x32/7)/6-16/6

 

y=8/7

 

so your parabola will be tangen to the point (32/7,8/7)

 

your equation will be:  y=(x-32/7(to shift it 32/7 right through x-axes))^2+8/7(to shift it 8/7 through y-axes) 

 

\(y=\left(x-\frac{32}{7}\right)^2+\frac{8}{7}\)

 

and not to pass through any other lines one of your equations will be:

\(y=1000\left(x-\frac{32}{7}\right)^2+\frac{8}{7}\)

OR

\(y=-1000\left(x-\frac{32}{7}\right)^2+\frac{8}{7}\)

 

here is your graph:

https://www.desmos.com/calculator/ccbc0v61sy

Solveit  Jan 18, 2016
 #2
avatar+2493 
+5

OMG! I found the equation of parabola :(((

Solveit  Jan 18, 2016
 #3
avatar+90968 
+5

Your answer and your graphs are very impressive Solviet but I do not understand what you have done.

I do not understand the question either   sad

 

 

Write the equation of hyperbole which is tangent with the straights 5x-6y-16=0 and 13x-10y-48=0.

This appears to be translated from another language.

I think it might mean

 

Write the equation of hyperbola where theses straights lines    5x-6y-16=0 and 13x-10y-48=0.    are the asymptotes.

 

This is an interesting question ://

Melody  Jan 18, 2016
 #4
avatar
+5

There is no unique solution to this.

The general solution is

65x^2 + 60y^2 - 128xy + 448x + 448y + C = 0,

where C is a constant.

Draw some graphs using desmos to check this out.

-Bertie

Guest Jan 19, 2016
 #5
avatar+90968 
+5

Thanks Bertie, it is really good to see you. :)

 

I could see that there was no unique solution but I did not know how to work out the general solution.

Can you show me how to do it?

 

I am not at all good with hyperbolas  sad

 

By the way, do you use Firefox as your browser?

and

Have you tried logging on using a different browser?

Melody  Jan 19, 2016
 #6
avatar
+10

Hi Melody

Here's the method I used, (though later I realised that there was something a whole lot quicker), and btw, there's a typo in the solution I gave earlier, the x term should be negative, -448x, apologies.

Just the framework, I'll leave you to fill in the algebra if you wish.

Start by shifting the intersection of the two lines to the origin,

X = x - 32/7, Y = y - 8/7.

The new lines are Y = 5X/6 and Y = 13X/10.

Assume that the equation of the hyperbola is Y^2 + aX^2 + bXY + c = 0.

(Can divide through by the coefficient of the Y^2 term so that it appears as 1 in the equation).

We need to see what happens to this as X and Y go to infinity.

Divide throughout by X^2,

(Y/X)^2 + b(Y/X) + a + c/X^2 = 0.

As X and Y go to infinity, this becomes (Y/X)^2 + b(Y/X) + a = 0, and that should give us the equations of the two asymptotes.

Using the usual formula for the solution of a quadratic,

(Y/X) = (-b +- sqrt(b^2 - 4a)) / 2.

The plus sign should give us the larger of the two gradients 13/10 and the negative sign the smaller, 5/6.

Solve for a and b.      a = 13/12, b = -32/15.

Now substitute into the earlier equations.

That gets you the equation I posted earlier (except for the typo I mentioned).

It occured to me when I'd been through this that I could have arrived at the result simply by multiplying the equations of the two lines together,

(5x - 6y - 16)(13x - 10y - 48) = 0,

though the meaning of the constant term 768 is interesting.

 

- Bertie

Guest Jan 19, 2016
 #7
avatar+2493 
+5

i found tangent parabola to the   5x-6y-16=0 and 13x-10y-48=0 but it is asking for hyperbole :/

Solveit  Jan 20, 2016
 #8
avatar+90968 
0

Thanks Bertie, 

I have only looked at your post superficially. I look forward to finding the time to study it and fill in the gaps. :)

CPhill told me privately that he was too :)

I have not finished with this one yet  ://     surprise

Melody  Jan 21, 2016
 #9
avatar+78551 
+5

Thanks for that answer, Bertie.....I went through the algebra as you suggested.......that's an interesting result to note that the product of both lines gives us the rotated conic equation.....I suspect that this might just be a coincidence, albeit a pretty fascinating one ....

 

Two questions......

 

The normal rotated conic usually contains an x and a y term, but I noticed that you omit them in your assumed equation........why is this???......

 

Also.....

 

Why do we let X and Y approach infinity??  ....Is this just to eliminate the c/X^2  term so we can solve the quadratic that follows??  I have seen Alan use this technique in solving another hyperbola problem......I was never able to wrap my head around it.....!!!!

 

All in all....this is a  really neat problem......

 

BTW......graphing this in Desmos just produces a pair of intersecting lines  .....https://www.desmos.com/calculator/quxgucllfk     ....????

 

 

cool cool cool

CPhill  Jan 21, 2016
 #10
avatar+26322 
+10

Here is WolframAlpha's answer:

 

hyperbola

Alan  Jan 21, 2016
 #11
avatar+18712 
+15

Hallo Bertie,

i have finished your wonderful solution.

 

\(\begin{array}{rcll} \text{I set slope line 1: } ~ m_1 = \frac{13}{10} \\ \text{and slope line 2: } ~ m_2 = \frac{5}{6} \\ \text{and your equation: }~Y^2 + aX^2 + bXY + c &=& 0. \qquad | \quad X = x - \frac{32}{7},~ Y = y - \frac87\\ \text{and your equations of the two asymptotes: }~(\frac{Y}{X})^2 + b\cdot\frac{Y}{X} + a &=& 0\\ \text{and with solution of a quadratic: }~\frac{Y}{X} &=& \frac12 \cdot (-b \pm \sqrt{b^2 - 4a})\\ \text{we can find a and b: } \end{array}\)

 

\(\begin{array}{rcll} \frac{Y}{X} &=& \frac12 \cdot (-b \pm \sqrt{b^2 - 4a}) \\ 2\cdot \frac{Y}{X}+b &=& \pm \sqrt{b^2 - 4a} \qquad | \qquad \text{(square both sides)} \\ (2\cdot \frac{Y}{X}+b)^2 &=& b^2 - 4a \\ 4\cdot (\frac{Y}{X})^2 +4\cdot\frac{Y}{X}\cdot b +b^2 &=& b^2 - 4a \\ 4\cdot (\frac{Y}{X})^2 +4\cdot\frac{Y}{X}\cdot b &=& - 4a \\ (\frac{Y}{X})^2 + \frac{Y}{X}b &=& -a \qquad | \qquad \text{we set }~ \frac{Y}{X}=m_1 \text{ and }\frac{Y}{X}=m_2\\ m_1^2+m_1b =&-a& =m_2^2+m_2b \\ b\cdot (m_1-m_2) &=& m_2^2-m_1^2 \\ b\cdot (m_1-m_2) &=&(m_2-m_1)(m_1+m_2) \\ b\cdot (m_1-m_2) &=&-(m_1-m_2)(m_1+m_2) \\ \mathbf{b} & \mathbf{=} & \mathbf{-(m_1+m_2)} = - (\frac{13}{10}+\frac{5}{6})=-\frac{32}{15} \\\\ m_1^2+m_1b &=& -a \\ m_2^2+m_2b &=& -a \\ b&=& \frac{-a-m_1^2}{m_1} = \frac{-a-m_2^2}{m_2} \\ m_2( -a-m_1^2 ) &=& m_1(-a-m_2^2 )\\ \cdots \\ a\cdot (m_1-m_2) &=& m_1\cdot m_2\cdot (m_1-m_2) \\ \mathbf{a} & \mathbf{=} & \mathbf{m_1\cdot m_2} = \frac{13}{10}\cdot \frac{5}{6} = \frac{13}{12}\\ \end{array}\)

 

\(\begin{array}{lcll} \text{your equation is now: } \end{array}\\ \begin{array}{rcll} Y^2 + m_1\cdot m_2\cdot X^2 - (m_1+m_2)\cdot XY + c &=& 0. \qquad | \quad X = x - \frac{32}{7},~ Y = y - \frac87\\ \end{array}\\ \begin{array}{lcll} \text{or: } \end{array}\\ \begin{array}{rcll} \mathbf{(y-\frac87)^2 + \frac{13}{10}\cdot \frac{5}{6} \cdot (x-\frac{32}{7})^2 - (\frac{13}{10}+\frac{5}{6})\cdot (x-\frac{32}{7}) \cdot (y-\frac87)+c }&\mathbf{=}& \mathbf{0} \end{array}\)

 

laugh

heureka  Jan 21, 2016
 #12
avatar
+10

Nicely done Heureka, but there is a quicker way.

Simply say that

(b + sqrt(b^2 - 4a))/ 2 = 13/10 so that  b + sqrt(b^2 - 4a)  = 13/5,

and

(b - sqrt(b^2 - 4a))/ 2 = 5/6 so that  b - sqrt(b^2 - 4a) = 5/3.

Add them, the sq roots cancel and you have the value of b directly.

Subtract one from the other, the b's cancel and you can square to find the value of a.

Guest Jan 21, 2016
 #13
avatar
+10

Hi Chris

 

The equation of an hyperbola centered at the origin with the x and y axes as its axes of symmetry can be written as

x^2/a^2 - y^2/b^2 = 1.

Introduce a rotation about the origin and that will produce an xy term.

Shift the centre away from the origin and that will give rise to x and y terms.

(A shift away from the origin, without a rotation, will give rise to x and y terms but no xy term).

 

 The two lines intersect at the point (32/7, 8/7) and this will be the centre of the hyperbola. The transformation X = x - 32/7, Y = y - 8/7, sets up new co-ordinate axes with origin at that point so that the equation of the hyperbola will not contain X and Y terms.

 

The asymptotes are the lines to which the  hyperbola tends as we move further and further away from the origin, that is, as X and Y tend to infinity. Dividing throughout by X^2 allows us to let X and Y tend to infinity, (on the assumption that the ratio Y/X remains finite).

 

If you multiply the equations of the lines together, you get

60y^2 + 65x^2 - 128xy - 448x +448y + 768 = 0,

and if you graph this, you get, as you should expect, a pair of straight lines, the lines from which the equation was derived.

Now change the value of the constant.

What you should see, is that if if you take the constant to be less that 768, 760 say or smaller, you will get a hyperbola which lies 'outside' the two straight lines, (and by that I mean that it occupies the regions for which the angle between the lines is obtuse).

If on the other hand you take values for the constant greater than 768 you should find that the resulting hyperbola lies between the two lines, ( the regions for which the angle between them is acute).

The fact that the equation of the hyperbola can be derived in this way isn't a coincidence, it should be true in most cases.

(I've just started to think about x = 0, y = 0 which should give rise to xy = 1 as an exception !).

 

- Bertie

Guest Jan 21, 2016
 #14
avatar+78551 
0

Thanks, Bertie, for that explanation.......very well done....!!!!!

 

 

cool cool cool

CPhill  Jan 21, 2016
 #15
avatar+90968 
+5

Thanks Solveit, Chis, Alan, Heureka and Bertie, :)

 

This one should go up in our Hall of Fame !

 

Lots to study and digest here.   

 

I always like that you are in the wings Bertie,  Thank you  laugh

 

(I assume the other guest post was also you )

Melody  Jan 21, 2016
 #16
avatar+18712 
+25
Best Answer

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

heureka  Jan 22, 2016
 #17
avatar+90968 
0

Thanks Heureka,    laugh

 

We really do need a HALL OF FAME  and this question needs to be in it !!!

Melody  Jan 25, 2016

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