Given a formula of hyperbola in standard form find foci, asymptotes, center vertices

Learn how to graph hyperbolas. To graph a hyperbola from the equation, we first express the equation in the standard form, that is in the form: (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1 for horizontal hyperbola or (y - k)^2 / a^2 - (x - h)^2 / b^2 = 1 for vertical hyperbola.
Next, we identify the characteristics of the given hyperbola. 'a' (the distance from the center to the vertices) is the square root of the first denominator and 'b' (the distance from the center to the covertices) is the square root of the second denominator. 'c' (the distance from the center to the foci) is obtained by taking the square root of the sum of a^2 and b^2. Using these characteristics of the hyperbola, we can graph the asymptotes of the hyperbola and hence graph the hyperbola.
Note that a hyperbola is vertical when it is facing up and down and is horizontal when it is facing right and left.
#conicsections #hyperbolaconicsections
#conicsections #hyperbolaconicsections
#conicsections #hyperbolaconicsections

[@eranilde7359]

Wonderful explanation... You have really helped me understand vertical and horizontal hyperbolas

[@SAMMINATIFF]

Thankyou sir you helped me such a big thanks to you sir I am in 11 th year this help me a lot and in which year you are teaching this conic section love from india

[@brianmclogan]

you are very welcome! I have been teaching conic sections for the past 6 years

[@evanmurphy6601]

Brian McLogan major help night before the final great stuff 👌

[@evananim6984]

THANK YOU VERY MUCH SIR, THIS REALLY HELPS A LOT🙇❤️

[@mikyllaguigayoma8241]

Sir, good morning. I just want to clarify the asymptote part. In our module, the formula for finding the asymptote when the center is at (0,0) is y=b/a, but in your formula, the a is in the numerator and the b is in the denominator. That is why I'm so confused on that part. Is y=b/a and y=a/b just the same? Also, when talking about finding the equation of the asymptotes, is a² always the one in the denominator of y² and b² is always the denominator of x²? Huhuhuuhuhuhu please read my questions. I'm really confused right now.

[@mikyllaguigayoma8241]

Ahhhhhhhh I get it now. After watching your videos related to this topic, I finally understand.If x² is the numerator in the first fraction, the formula will be y=k±(b/a)x(x-h) If y² comes first, the formula is y=k±(a/b)x (x-h). Omgggggg I finally get it😭😭💕💕 Thankk youuuuu very much😙😭💕💕

[@brianmclogan]

So the asymptote equation depends on if the transverse axis is vertical or horizontal. And no a^2 is not always under x^2 but listed first

[@mikyllaguigayoma8241]

@@brianmclogan yes, sir!!! Thank you very much!! WAHHHHHHHHH YOU REPLIED OMGGGGG THANK YOU😭😭🥺💖💖

[@queenofscore]

Thank you so much teacher you really help me.

[@SAMMINATIFF]

You are like god sir hats of you

[@brianmclogan]

happy to help

[@rodrigosanchez1969]

Thanks, professor. You make hyperbola easier

[@matthewgage1054]

*Sees man carving a miniature sculpture while riding on a longboard* (edit): sorry wrong tab

[@rheadomingo5797]

how did you get the 5 in vertices

[@samanthanicoleasilo9633]

From the value of a² = 25, and square both sides which gives you a = 5

[@bryonsegunto8901]

Sir what's the formula for directrices?

[@brianmclogan]

That is not required in my class so I do not have anything on them

[@ladyannebaldon8907]

sir lang malakas hehehe

[@doperae]

HAHAHHAHAHA sir Brian lang malakas

[@evimelsarmiento5531]

Sir. Whats the general equation of hyperbola?

[@shenyang6007]

Horizontal: [(X-h)^2/a^2]-[(Y-k)^2/b^2)]=1 Vertical: [(Y-k)^2/a^2]-[(X-h)^2/b^2)]=1 (h, k) is the center coordinate a is distance from vertex to center b is distance of minor axis c is distance from foci to center

[@AAAAAANICKKKKKK]

If "x² - y²/25 = 1" Then a = 1, b = 5 then asymptote is "y = (±5/1)(x)" or "y = ±5x"? I can't understand my teacher in online class :

[@kfjfkeofitorhf9520]

A^2==25===(5) B^2==81===(9) C===106===(106)

[@tobeythegoat6735]

What a legend 💛

[@sontzyguxx468]

Stem pa

[@shaenism_]

thankyou!!🥺

[@brianmclogan]

You are very welcome!!

[@bigjimmy3792]

Holy snore cut to the chase and tell me how to do the damn equation

[@jeffreybals4430]

🤢

[@laarnemaecaballero8479]

THANK YOU VERY MUCH SIR, THIS REALLY HELPS A LOT🙇❤️