My savior... learned more in 7 minutes than 2 full classes in a college course... really goes to show you the difference when you have a competent teacher. Thanks a bunch!
this is random but I love, love, love your thumbnails lmao. its easy to identify. thank you soooo much you've been helping me more than my actual teacher LMAOAOO
Will you please be my teacher??!!! I was crying in a corner while my math book pointed and laughed at me all night! (at least thats what it felt like lol) I have learned so much from you, you rock! Thanks !!
+Virginia Klotz you are very welcome! Will be uploading all year long so please keep in touch and let me know if there is anything else I can do for you
Parabola: Vertex is the midpoint between the directrix and the foci, there is no "center". Ellipse: center is midpoint between two foci. Hyperbola: Center is midpoint between two vertices. Distances and "C": Parabola: Distance between foci and vertex= C Ellipse (major/minor axis): Distance between center and foci= C Hyperbola: a- distance between center and vertices c- distance between center and foci Equations: Parabola @ origin (Vertex (0,0) ). x^2=4cy focus (o,c/p) *If c is positive graph opens up, if C is negative graph opens down* y axis is axis of symmetry or x=0 or y^2=4cx focus (c,0) and *If c is positive graph opens right, if C is negative graph opens left* x axis is axis of symmetry or y=0. Center (h,k) (x-h)^2=4c(y-k) Focus (h,k+c) Directrix: y=k-c or (y-k)2=4c(x-h) Focus: (h+c,k) Directrix x=h-c Ellipse @ Origin Center (0,0) x^2/a^2+ y^2/b^2=1 *major axis on the x^2, a^2 always goes on the major axis, in this case x^2* vertices (-a,0),(a,0) y-intercepts (0,-b), (0,b) Foci: (+/-c,0) and C^2=a^2-b^2 or Major axis on y^2 x^2/b^2+y^2/a^2=1 vertices (o,+/-a) x-int (+/-b,0) Foci: (0,+/-C) @ (h,k) (x-h)^2/a^2+(y-k)^2/b^2=1 *Major axis is x^2* 0r (x-h)^2/b^2+(y-k)^2/a^2=1 *Major axis is y^2* Hyperbola: @ center (0,0) X or Y will be first if that is the transverse axis x^2/a^2-y^2/b^2=1 *graph never crosses the y-axis* asymptote* oblique asymptote: y=+/- b/ax or y^2/a^2-x^2/b^2=1 *graph never cross x-axis* c^2=a^2+b^2 with center at h,k (x-h)^2/a^2-(y-k)2/b^2=1 vertices: (h-a,k),(h+a,k) foci: (h-c,k),(h+c,k) or (y-k)^2/a^2-(x-h)^2/b^2=1 vertices: (h,k-a),(h,k+a) Foci: (h,k-c) (h,k+c) where c^2=a^2+b^2 I hope this helps guys!
Foci of (+/- 8, 0), center at origin, major axis of 18. Would that cause a and b to be equal, therefore create a circle? I got ((x-0)^2/81 + (y-0)^2/81)= 1 Disregard. Answer is: (x^2/81 + y^2/17)= 1
Thank you very much sir, I'm just following you to solve mine then I'm so surprised that I got the right answer......... You really do very well sir. Thanks a lot.
OMG this is what teaching looks like. all my teacher does is read notes. even i could do her job, since she doesn't explain crap, just reads textbooks,
In this case it didn't matter because he got lucky but in a different problem the B² would be incorrect
yes the correct formula is c^2=a^2-b^2
instablaster.
My savior... learned more in 7 minutes than 2 full classes in a college course... really goes to show you the difference when you have a competent teacher. Thanks a bunch!
thank you! I really appreciate it and am very happy to help you out
this is random but I love, love, love your thumbnails lmao. its easy to identify. thank you soooo much you've been helping me more than my actual teacher LMAOAOO
Sir, you're a life saver! Thank you so much!
(2,1) is the center not the vertex.
correct, I misspeak all the time, that you for catching it
Correct
I wish I'm one of your students, your so great on teaching.
Super thank you to you sir! I always go to your channel whenever i have assignments!
This is awesome. I have an exam tomorrow huhuhu life saver!
best of luck!
Thank you sir! Your content is very useful this time of pandemic. Thank you sir!
Happy to help!!
this guy makes math possible again. my teacher doesnt even teach :(
Will you please be my teacher??!!! I was crying in a corner while my math book pointed and laughed at me all night! (at least thats what it felt like lol) I have learned so much from you, you rock! Thanks !!
+Virginia Klotz you are very welcome! Will be uploading all year long so please keep in touch and let me know if there is anything else I can do for you
no prob
Parabola: Vertex is the midpoint between the directrix and the foci, there is no "center".
Ellipse: center is midpoint between two foci.
Hyperbola: Center is midpoint between two vertices.
Distances and "C":
Parabola: Distance between foci and vertex= C
Ellipse (major/minor axis): Distance between center and foci= C
Hyperbola: a- distance between center and vertices
c- distance between center and foci
Equations:
Parabola @ origin (Vertex (0,0) ).
x^2=4cy focus (o,c/p) *If c is positive graph opens up, if C is negative graph opens down* y axis is axis of symmetry or x=0
or
y^2=4cx focus (c,0) and *If c is positive graph opens right, if C is negative graph opens left* x axis is axis of symmetry or y=0.
Center (h,k)
(x-h)^2=4c(y-k) Focus (h,k+c) Directrix: y=k-c
or
(y-k)2=4c(x-h) Focus: (h+c,k) Directrix x=h-c
Ellipse @ Origin Center (0,0)
x^2/a^2+ y^2/b^2=1 *major axis on the x^2, a^2 always goes on the major axis, in this case x^2*
vertices (-a,0),(a,0) y-intercepts (0,-b), (0,b) Foci: (+/-c,0) and C^2=a^2-b^2
or
Major axis on y^2
x^2/b^2+y^2/a^2=1
vertices (o,+/-a) x-int (+/-b,0) Foci: (0,+/-C)
@ (h,k)
(x-h)^2/a^2+(y-k)^2/b^2=1 *Major axis is x^2*
0r
(x-h)^2/b^2+(y-k)^2/a^2=1 *Major axis is y^2*
Hyperbola: @ center (0,0)
X or Y will be first if that is the transverse axis
x^2/a^2-y^2/b^2=1 *graph never crosses the y-axis* asymptote*
oblique asymptote: y=+/- b/ax
or
y^2/a^2-x^2/b^2=1 *graph never cross x-axis*
c^2=a^2+b^2
with center at h,k
(x-h)^2/a^2-(y-k)2/b^2=1
vertices: (h-a,k),(h+a,k) foci: (h-c,k),(h+c,k)
or
(y-k)^2/a^2-(x-h)^2/b^2=1
vertices: (h,k-a),(h,k+a) Foci: (h,k-c) (h,k+c) where c^2=a^2+b^2
I hope this helps guys!
thanks for sharing
Foci of (+/- 8, 0), center at origin, major axis of 18.
Would that cause a and b to be equal, therefore create a circle?
I got ((x-0)^2/81 + (y-0)^2/81)= 1
Disregard. Answer is: (x^2/81 + y^2/17)= 1
It's helpful 👍
Thank you very much sir, I'm just following you to solve mine then I'm so surprised that I got the right answer......... You really do very well sir. Thanks a lot.
Clear explanation; easy to understand! Thanks!!!
happy to be able to help
thank you so much. My Pre-cal teacher is horrible, and you are the only reason that I am scraping by with a B.
+Lilith Jackson happy to help! keep up the hard work
Thank you so much, been watching your videos for years and it truly helped me a lot.
You are very welcome!!
Ah!!!this saved me a lot sir thank you so so much
OMG this is what teaching looks like. all my teacher does is read notes. even i could do her job, since she doesn't explain crap, just reads textbooks,
How can i find equation with foci and minor axis given?
Thanks bro
Thank you!!! My teacher doesnt do examples so i have to go to your channel lol
FourteyNiner Fan happy to be there for you! just let me know if there is anything I can do for you.
How do you find the equation of the hyperbola if the given is only the two foci and a difference in distance from any point to the foci?
Find equation of hyperbola if vertices are (2,4) ,(10,4) and foci are 10 units apart
I have many examples just like that find a c and the center from that information then plug it into the equation
Great explanation
Thank you!
The equation to find b^2 is c^2= a^2 minus b^2
correct, thank you I pinned your comment
how should i do iit, if the foci is not perfect square? or the givin foci is 4√
2
how do i get the b if the center is at 0? it'll be a circle if it has the same number of the a.
Matik y²
cool! extremely helpful! thanks so much for this video!
very welcome!
:-D
how do u find the center if u only have 1 focus?
When i use the midpoint formula i got the center (2,0)
If it was (2,0) the point would not be on the major axis
life saver
ty sir
you are very welcome!
"2 foci's" = 2 foci. 1 focus •x = x foci
Sir what if the foci is (-4±✓15,3)
And the major axis is 10 units long
Same process.
Any indian here 😀
Thahhh DuHHH :-) Just like that :-)
Thank you for sharing :-)
you are very welcome!
how do you even find B?
C^2=a^2-b^2
would be better to watch the video if it didn't have any errors.
6 major
Then the a² 9 ???
I think is 36 right
Isnt he incorrect? A foci should be greater than the vertice... its always outside the vertices... in this case its inside....??
"Horizontal major axis of length 8, center at the origin and passes through 2,5
"
how to get the b and c?
Hahaha I saw this problem in my quiz AHAHAHAHAHA
center not vertex
can you make it in the vertwx of the origin