The definition of hyperbola, is a graph in which at any point, the difference in the distances from that point to 2 fixed points (the foci) is CONSTANT. and From this definition and by how the graph looks if you were to make select the point (a,0) and calculate the difference in distances to the foci (c,0) and (-c,0) you will get D1 - D2 = a- c + a - (-c) = 2a. Thus we know that at point (a,0) D1 - D2 = 2a, by definition D1 - D2 is constant, thus it equals 2a at ALL points in a hypbola
Perfect! Even the most tedious derivations are important in order to keep learning maths. It is not that you have to remember exactly how to get it. It is important to know that there is a way to get it and important to have walked that way at least once.
CORRECTION: i made a little mistake, at point (a,0) the difference in distances would be D1 - D2 = (c - a) - (c+a) = -2a. same answer though but negative. if you select D2 - D1 you would get positive, and hence the +/-2a UPDATE: Please my newer better more updated video on the Hyperbola! ruclips.net/video/UBIHovXNV9U/видео.html
yeah OR when you hit x^2((c^2)/(a^2)-1)-x^2-y^2 = c^2-a^2 you just multiply the first term by a^2 (top and bottom) and then divide the whole thing by the right side and you get x^2/a^2 - y^2/(c^2-a^2) = 1
Hi Niraj, thanks for asking. I defined b^2 = c^2 - a^2. Thus rearranging we get c^2 = a^2 + b^2. I did this so that we can write the parabola more neatly.
I chose b^2 = c^2 - a^2 because it makes the formula look better. That is the only reason. If you derive the formula without choosing this b parameter than what you get is: y^2/a^2 - x^2/(c^2-a^2) = 1. Since both c and a are constants than we can simply choose b^2 = c^2-a^2 (which is also a constant) so that the equation looks better: y^2/a^2 - x^2/b^2 = 1. I hope this helps!
Hi Jack, thanks for asking. I defined b^2 = c^2 - a^2. Thus rearranging we get c^2 = a^2 + b^2. I did this so that we can write the parabola more neatly. The fact that it is also the pythagoras theorem is just a coincidence (or more accurately different mathematical concepts usually make their way in other seemingly unrelated math concepts).
Thanks for asking. I made a better more updated video here: ruclips.net/video/UBIHovXNV9U/видео.html Remember that distances D1 and D2 are measured from the foci and NOT the origin.
The definition of hyperbola, is a graph in which at any point, the difference in the distances from that point to 2 fixed points (the foci) is CONSTANT. and From this definition and by how the graph looks if you were to make select the point (a,0) and calculate the difference in distances to the foci (c,0) and (-c,0) you will get D1 - D2 = a- c + a - (-c) = 2a. Thus we know that at point (a,0) D1 - D2 = 2a, by definition D1 - D2 is constant, thus it equals 2a at ALL points in a hypbola
Perfect! Even the most tedious derivations are important in order to keep learning maths. It is not that you have to remember exactly how to get it. It is important to know that there is a way to get it and important to have walked that way at least once.
i think the singular form of "vertices" is "vertex" not "vertice"............... ??? .-.
thank you so much it all makes perfect sense now , i missed this lesson.
you're a good teacher
The Pythagoras theorem is used in setting up the equation of the distances, i.e. D^2 = y^2 + (x-c)^2
why is D1 - D2 = 2a. is that according to the definition of a hyperbola? or can u derive it from somewhere..? thx!
CORRECTION: i made a little mistake, at point (a,0) the difference in distances would be D1 - D2 = (c - a) - (c+a) = -2a. same answer though but negative. if you select D2 - D1 you would get positive, and hence the +/-2a
UPDATE: Please my newer better more updated video on the Hyperbola! ruclips.net/video/UBIHovXNV9U/видео.html
c^2-a^2/b^2 is 1, not b^2 (24:46)
yeah OR when you hit x^2((c^2)/(a^2)-1)-x^2-y^2 = c^2-a^2 you just multiply the first term by a^2 (top and bottom) and then divide the whole thing by the right side and you get x^2/a^2 - y^2/(c^2-a^2) = 1
plz how to proof
asymptose equation of hyperbola
Thanks for asking. I will see if I can do a video on this.
but i think this solution works only if we consider D1 to be bigger.
computer lover yep
where does pythagoras theorem come in too??
thank you!
you're welcome! :)
c^2 = a^2 + b^2... how?????...please explain
Hi Niraj, thanks for asking. I defined b^2 = c^2 - a^2. Thus rearranging we get c^2 = a^2 + b^2. I did this so that we can write the parabola more neatly.
I know , what I don't understand is from where it comes??? that is c^2 = b^2 - a^2 ,,,,,,,this is my problem ....
I chose b^2 = c^2 - a^2 because it makes the formula look better. That is the only reason. If you derive the formula without choosing this b parameter than what you get is: y^2/a^2 - x^2/(c^2-a^2) = 1. Since both c and a are constants than we can simply choose b^2 = c^2-a^2 (which is also a constant) so that the equation looks better: y^2/a^2 - x^2/b^2 = 1. I hope this helps!
why was the relationship between a,b and c defined by pythagoras?
Hi Jack, thanks for asking. I defined b^2 = c^2 - a^2. Thus rearranging we get c^2 = a^2 + b^2. I did this so that we can write the parabola more neatly. The fact that it is also the pythagoras theorem is just a coincidence (or more accurately different mathematical concepts usually make their way in other seemingly unrelated math concepts).
thank u so much this helped me a lot!! :)
Jocelyn Hiranyajinda really？
please slove the eccentricity value 1.5
thanks! :)
Very informative, but I fell asleep. I'm sorry :L
oh ok that was good but I was slow to pick up the expansions
Thanks for asking. I made a better more updated video here: ruclips.net/video/UBIHovXNV9U/видео.html
Remember that distances D1 and D2 are measured from the foci and NOT the origin.
such a long vedio