Almost one third of the course to this whole Mathematics series, want to confess... had I understood all this in this way 15 years back when it was taught to me in my school, I would have been a different person today. Alas! teachers took no interest in making average students comfortable in Maths classes. They were happy with so called brighter ones in class. Undoing the harm done that time by understanding all this now from you. Lucky that stumbled upon your account. Kudos to you @ProfessorDaveExplains. 🤗
Your comment sums up exactly how I feel. Every single word is spot on, especially where you stated that "...teachers took no interest in making average students comfortable...:
@@kentishbrigant2053 Me too! Maths classes would start with a concept then build upon it. If you didn't get the concept quickly, you'd play catch up for the rest of the lesson and likely the whole term. If I'd had RUclips and Prof Dave's videos maths would have actually been enjoyable rather than something to be terrified of.
@srishti_chaurasia_09 Former "brighter" student here. I honestly feel this in a way. Although, for _my_ part it was because I initially didn't have to study, so by the time I _did_ have to study (because you *_never_* can reach an optimal amount of knowledge without _at least _*_some_* studying), I didn't even know it, _let alone_ how I'd go about it. These systems kinda give _everyone_ the short end of the stick. 😅
The standard form equation for an ellipse centered at $(h,k)$ is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $a$ is the distance from the center to a vertex along the major axis and $b$ is the distance from the center to a co-vertex along the minor axis. In this case, the center is given as $(-2,4)$, so $h=-2$ and $k=4$. Additionally, we know the foci and vertices, which provide key information about the dimensions of the ellipse: - The distance between the center and each focus is $c$, so $c = 5 - 4 = 1$. - The distance between the center and each vertex along the major axis is $a$, so $a = 9 - 4 = 5$. - The distance between the center and each co-vertex along the minor axis is $b$, so $b = 1$. Now we have all the information we need to write the equation: \frac{(x+2)^2}{5^2} + \frac{(y-4)^2}{1^2} = 1 This is the equation in standard form of an ellipse with center (-2,4), foci at (-2,3) and (-2,5), and vertices at (-2,-1) and (-2,9)
brothers in the comprehension part 2nd question if anyone have problem just draw graph it is actually easy!!!!!!!!!!!! c= distance from center to focus(foci) , it is 1 from both x and y axis when you draw graph you can see ; a=distance from center to vertex which is 5 remember center is not (0,0) rather (-2,4) so count everything from that position. in the last part, since y axis tall ,it is taller ellipse, that is why the formula will be, (x-h)²/b² + (y-k)²/a² = 1 hope this helps!
When we are looking at the foci and the vertices to be plugged into c^2= a^2 + b^2, is there a particular order in which the points need to be plugged in? For example, does the x value of the first or second parenthesis of the foci go in for a^2 or not? The same question goes for the vertices with c^2. I would appreciate the help. Thanks
youtube please put his conic section videos up first when I search the topic, not the one hour long videos! I had to search for your channel specifically I feel bad for all the students who haven't found you yet : (
If you could add how to derive the equation of ellipse, this video would be perfect. And all function transformation could apply for ellipse if we only take one fourth of ellipse as function. Thus, all ellipse is applicable for transformation.
I had not understood this tutorial so i just skipped so i could come back later, just came back right after watching the hyperbola one! (the hyperbola and the ellipse concepts are very similar)
I think I have found a tiny mistake. The second equation (in checking comprehension) i think 25 and 24 have been swapped in the final result. @5:53 Awesome channel, btw.
@@malekblr8823 The foci share the same x value (-2) so the ellipse has to be vertical. If the foci share the same y value the ellipse is horizontal. The foci always are points based on the axis the ellipse stretches. (-3, 5) & (-3, 23) --> vertical ellipse (-6, 4) & (12, 4) --> horizontal ellipse
1. For the ellipse with center (0, 0), foci at (-2, 0) and (2, 0), and vertices at (-4, 0) and (4, 0) First, let's find the distance between the foci. The distance between the foci is given by the formula: c = distance between foci = 2a Since the foci are located at (-2, 0) and (2, 0), we can see that the distance between them is 4 units. Therefore, c = 4. Next, let's find the value of a, which represents half the length of the major axis. The distance between the center (0, 0) and each vertex (-4, 0) or (4, 0) is equal to a. So, a = 4. Now, we can use the values of a and c to determine the equation of the ellipse. The general equation for an ellipse with its center at (0, 0) is: (x^2 / a^2) + (y^2 / b^2) = 1 Since the center is (0, 0), the equation becomes: (x^2 / 4^2) + (y^2 / b^2) = 1 Simplifying further, we have: (x^2 / 16) + (y^2 / b^2) = 1 To find the value of b, we can use the relationship between a, b, and c: c^2 = a^2 - b^2 Substituting the known values, we get: 4^2 = 4^2 - b^2 16 = 16 - b^2 b^2 = 0 This tells us that b = 0. However, an ellipse cannot have a major axis of length 8 (2a) and a minor axis of length 0. Thus, there seems to be an error in the given information. An ellipse cannot be formed with the provided foci and vertices.
Anyone else watching and seeing how the 2nd comprehension example is almost a proof for why every sphere is an ellipse? 😁 To clarify imagine said equation being (y - 4)²/25 + (x + 2)²/25 = 1 Multiply by 25 (y - 4)² + (x + 2)² = 25 And you got the standard form of a sphere 😉 (x - h)² + (y - k)² = r² Or based on the equation from before (x + 2)² + (y - 4)² = 25 So remember: if your ellipse equation has a = b it's actually a sphere, not only an ellipse. Also pretty obvious if you look at it as a graph ( 1:40 ) ofc.
@@ProfessorDaveExplains I'm currently binge watching the whole maths playlist (well, almost, started with the algebra part) and it's kinda mind blowing how often I sit here thinking "ohhhh...this makes perfect sense" with things that just come to mind now with all the knowledge and wisdom I accumulated since I finished school ~20yrs ago because they might not even be explained or mentioned in the video. For me your videos are a major pillar of science communication and education on the whole internet and I'm very much looking forward to revisiting those videos when my kid might struggle with some of the covered topics at school. Thank you VERY MUCH for all the work you put into this channel Dave! You are doing a great job, please keep it up! (I mean, I kinda know you will anyway) PS: Why am I giggling like a 12yr old girl at a boyband concert? The answer is probably found in one of your psychology videos - but I'm pretty sure I already know it😁
c represents the distance between the center and any one of the foci. Coordinate of the center: (-2, 4), coordinate of the first focus: (-2, 3). Distance between these two points = 1. So, c = 1. 1*1 = 1 as well, therefore c squared equals one
sorry im not quite understand why the standard form must contains 1,and why 1 is so important, please answer my question,thank you, btw great explaining
No, because the center has a value of input and output but I think what you meant to ask is if 25 can be placed under the x but no you just take 25 from y and substitute it with the a value
A is the distance between the vertex and the center, so you take (9-(-1))/2 this is because this ellipse is longer than it is wide, therefore, the a is on the y axis. We divide by 2 because a is equal to half of the value of the major axis.
An oval is any flat, closed, continuously curved shape, where the curvature is always directed inward. An ellipse is a special case of an oval, that has a precise definition in mathematics. If you scale a circle in one direction, while the other dimension remains unchanged, that is an ellipse. If you slice the cone on the diagonal that is not as steep as the side of the cone, to make the conic section, that is an ellipse. If its shape is defined by the sum of the distances between two foci being constant, such that you can use two tacks and a string to draw it, that is an ellipse. All ellipses are ovals, but not all ovals are ellipses.
Plot the given pairs(vertices, and foci) , center first on the coordinate plane then count the vertices and foci starting from the center if you get the same value from counting both sides then that is your b²=
If the foci share the same x value (as with the 2nd comprehension example) the ellipse is vertical, so "a" has to be with y. If the foci share the same y value (as with the 1st comprehension example) the ellipse is horizontal, so "a" has to be with x. So if you want to know if the ellipse stretches vertical or horizontal all you need to know are the 2 foci, no need for vertices. The axis the foci share is the axis the ellipse stretches.
@@Teiwaz111I appreciate this clarification I find it more faster to determine whether the ellipse is horizontal or vertical using your instructions before I saw this I would first plot the center and count the units of x and y axis from left to right 😂
Does it matter if we express an equation of an ellipse as (x^2)/(a^2) + (y^2)/(b^2) = 1 OR (y^2)/(b^2) + (x^2)/(a^2) = 1 ? (since at the end of the day, both expressions are commutative)
Almost one third of the course to this whole Mathematics series, want to confess... had I understood all this in this way 15 years back when it was taught to me in my school, I would have been a different person today. Alas! teachers took no interest in making average students comfortable in Maths classes. They were happy with so called brighter ones in class.
Undoing the harm done that time by understanding all this now from you. Lucky that stumbled upon your account. Kudos to you @ProfessorDaveExplains. 🤗
Your comment sums up exactly how I feel. Every single word is spot on, especially where you stated that "...teachers took no interest in making average students comfortable...:
@@kentishbrigant2053 Me too! Maths classes would start with a concept then build upon it. If you didn't get the concept quickly, you'd play catch up for the rest of the lesson and likely the whole term.
If I'd had RUclips and Prof Dave's videos maths would have actually been enjoyable rather than something to be terrified of.
@srishti_chaurasia_09
Former "brighter" student here. I honestly feel this in a way. Although, for _my_ part it was because I initially didn't have to study, so by the time I _did_ have to study (because you *_never_* can reach an optimal amount of knowledge without _at least _*_some_* studying), I didn't even know it, _let alone_ how I'd go about it.
These systems kinda give _everyone_ the short end of the stick. 😅
Excellent, concise breakdown
you're my lifesaver this week between chem and pre calc. me and my roomie
Thanks Prof Dave for explaining the topic very clearly! You have been very helpful! Keep up the good work!
You are a hero Professor Dave. Thanks so much for these.
you the best Prof ever
Thank you so much for uploading this!!
Excellent video ! Thanks professor Dave !
The standard form equation for an ellipse centered at $(h,k)$ is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $a$ is the distance from the center to a vertex along the major axis and $b$ is the distance from the center to a co-vertex along the minor axis.
In this case, the center is given as $(-2,4)$, so $h=-2$ and $k=4$. Additionally, we know the foci and vertices, which provide key information about the dimensions of the ellipse:
- The distance between the center and each focus is $c$, so $c = 5 - 4 = 1$.
- The distance between the center and each vertex along the major axis is $a$, so $a = 9 - 4 = 5$.
- The distance between the center and each co-vertex along the minor axis is $b$, so $b = 1$.
Now we have all the information we need to write the equation:
\frac{(x+2)^2}{5^2} + \frac{(y-4)^2}{1^2} = 1
This is the equation in standard form of an ellipse with center (-2,4), foci at (-2,3) and (-2,5), and vertices at (-2,-1) and (-2,9)
Great presentation!
very useful. Thanks for break it down
Beautiful.
brothers in the comprehension part 2nd question if anyone have problem just draw graph
it is actually easy!!!!!!!!!!!!
c= distance from center to focus(foci) , it is 1 from both x and y axis when you draw graph you can see ;
a=distance from center to vertex which is 5
remember center is not (0,0) rather (-2,4)
so count everything from that position.
in the last part, since y axis tall ,it is taller ellipse, that is why the formula will be, (x-h)²/b² + (y-k)²/a² = 1
hope this helps!
yo thanks bro you are goated
i really had hard time about understanding this but you explained to me very well thanks
@@Teenageexperience-nd4xz pleasure is mine
بارك الله فيكم وجزاكم الله خير الجزاء
This video is online gold.
how do we derive the standard form of the equation of ellipse?
Check mathematicsonline's video about this
@@jiminsbluemold 6 years too late buddy 🥲
@@Greentealiesel better late than never 😁
@@jiminsbluemold BETTER LATE THAN NEVER BETTER LATE THAN NEVER 📢📢😼😼
Bruh.... Imagine if harshsinghal replies
When we are looking at the foci and the vertices to be plugged into c^2= a^2 + b^2, is there a particular order in which the points need to be plugged in? For example, does the x value of the first or second parenthesis of the foci go in for a^2 or not? The same question goes for the vertices with c^2. I would appreciate the help. Thanks
Every time I see this shape I think of the PHP 5 logo.
youtube please put his conic section videos up first when I search the topic, not the one hour long videos!
I had to search for your channel specifically I feel bad for all the students who haven't found you yet : (
If you could add how to derive the equation of ellipse, this video would be perfect. And all function transformation could apply for ellipse if we only take one fourth of ellipse as function. Thus, all ellipse is applicable for transformation.
How do you derive it?
why c^2 = a^2 - b^2
Also wanted to know that
Pythagorean theorem
THANK YOU FOR THIS, I NEEDED THIS!!!!!
You’re my hero
Very helpful ❤️
Done this lesson.
I had not understood this tutorial so i just skipped so i could come back later, just came back right after watching the hyperbola one! (the hyperbola and the ellipse concepts are very similar)
Will you be explaining eccentricity in another video beyond telling us what the values are for various types of conic sections?
Thank you math Jesus
2:51 when do we use c^2 = a^2 - b^2 and c^2 = a^2 + b^2? You should've put that before introducing this equation.
+ is for the hyperbola
- is ellipse
Thank you soooooo much
I think I have found a tiny mistake.
The second equation (in checking comprehension) i think 25 and 24 have been swapped in the final result. @5:53
Awesome channel, btw.
hmm no i think it's correct, because the ellipse is vertical so the a term goes with the y
Thank you so much, I understand better :)
@@ProfessorDaveExplains but how did we knew that it was vertical
@@malekblr8823 The foci share the same x value (-2) so the ellipse has to be vertical.
If the foci share the same y value the ellipse is horizontal.
The foci always are points based on the axis the ellipse stretches.
(-3, 5) & (-3, 23) --> vertical ellipse
(-6, 4) & (12, 4) --> horizontal ellipse
@@malekblr8823 draw and checking
Can someone explain how to solve the second one?
1. For the ellipse with center (0, 0), foci at (-2, 0) and (2, 0), and vertices at (-4, 0) and (4, 0)
First, let's find the distance between the foci. The distance between the foci is given by the formula:
c = distance between foci = 2a
Since the foci are located at (-2, 0) and (2, 0), we can see that the distance between them is 4 units. Therefore, c = 4.
Next, let's find the value of a, which represents half the length of the major axis. The distance between the center (0, 0) and each vertex (-4, 0) or (4, 0) is equal to a.
So, a = 4.
Now, we can use the values of a and c to determine the equation of the ellipse. The general equation for an ellipse with its center at (0, 0) is:
(x^2 / a^2) + (y^2 / b^2) = 1
Since the center is (0, 0), the equation becomes:
(x^2 / 4^2) + (y^2 / b^2) = 1
Simplifying further, we have:
(x^2 / 16) + (y^2 / b^2) = 1
To find the value of b, we can use the relationship between a, b, and c:
c^2 = a^2 - b^2
Substituting the known values, we get:
4^2 = 4^2 - b^2
16 = 16 - b^2
b^2 = 0
This tells us that b = 0. However, an ellipse cannot have a major axis of length 8 (2a) and a minor axis of length 0. Thus, there seems to be an error in the given information. An ellipse cannot be formed with the provided foci and vertices.
valiant effort
c isn't the distance between the foci; rather it is the distance *from* the center *to* either of the foci. That means c is only 2 man.
C is the distance from the center between either of the foci
Anyone else watching and seeing how the 2nd comprehension example is almost a proof for why every sphere is an ellipse? 😁
To clarify imagine said equation being
(y - 4)²/25 + (x + 2)²/25 = 1
Multiply by 25
(y - 4)² + (x + 2)² = 25
And you got the standard form of a sphere 😉
(x - h)² + (y - k)² = r²
Or based on the equation from before
(x + 2)² + (y - 4)² = 25
So remember: if your ellipse equation has a = b it's actually a sphere, not only an ellipse.
Also pretty obvious if you look at it as a graph ( 1:40 ) ofc.
Right, a circle is an ellipse where the two foci coincide, so we call it something else, the center.
@@ProfessorDaveExplains I'm currently binge watching the whole maths playlist (well, almost, started with the algebra part) and it's kinda mind blowing how often I sit here thinking "ohhhh...this makes perfect sense" with things that just come to mind now with all the knowledge and wisdom I accumulated since I finished school ~20yrs ago because they might not even be explained or mentioned in the video.
For me your videos are a major pillar of science communication and education on the whole internet and I'm very much looking forward to revisiting those videos when my kid might struggle with some of the covered topics at school.
Thank you VERY MUCH for all the work you put into this channel Dave! You are doing a great job, please keep it up! (I mean, I kinda know you will anyway)
PS: Why am I giggling like a 12yr old girl at a boyband concert? The answer is probably found in one of your psychology videos - but I'm pretty sure I already know it😁
Great!
Sir can u please explain how c square =1 in the comprehension 2nd problem
c represents the distance between the center and any one of the foci. Coordinate of the center: (-2, 4), coordinate of the first focus: (-2, 3). Distance between these two points = 1. So, c = 1. 1*1 = 1 as well, therefore c squared equals one
@@romankozelov5078 Wow! Thank you Roman!
sorry im not quite understand why the standard form must contains 1,and why 1 is so important, please answer my question,thank you, btw great explaining
I believe that it originally derives from a trig identity!
in comprehension question #2, can it also be x^2/24 + y^2/25 = 1 like in 3:15 of the video?
No, because the center has a value of input and output but I think what you meant to ask is if 25 can be placed under the x but no you just take 25 from y and substitute it with the a value
The formula would be (x-h)²+(y-k)² if the center has a value
How did you solve the second one?
A is the distance between the vertex and the center, so you take (9-(-1))/2 this is because this ellipse is longer than it is wide, therefore, the a is on the y axis. We divide by 2 because a is equal to half of the value of the major axis.
Great ❤️
you are adjust the position of talent
3:50 this is multiplication though???????
Get the reciprocal first then proceed to multiplication or you can simply cross multiply
One second... how did ya get the value of a and c from (-1,0) & (1,0) and (-2,0) & (2,0) .
Please help
A is the distance between the vertex and center, so a = 4 because this ellipse is longer than it is wide.
What is the difference between an ellipse and an oval?
An oval does not have a precise mathematical definition.
An oval is any flat, closed, continuously curved shape, where the curvature is always directed inward.
An ellipse is a special case of an oval, that has a precise definition in mathematics. If you scale a circle in one direction, while the other dimension remains unchanged, that is an ellipse. If you slice the cone on the diagonal that is not as steep as the side of the cone, to make the conic section, that is an ellipse. If its shape is defined by the sum of the distances between two foci being constant, such that you can use two tacks and a string to draw it, that is an ellipse.
All ellipses are ovals, but not all ovals are ellipses.
ngl the 2nd question is hard, the jump in difficulty is big for me atleast. could use a bit more explaining in the other half
Plot the given pairs(vertices, and foci) , center first on the coordinate plane then count the vertices and foci starting from the center if you get the same value from counting both sides then that is your b²=
AHHHHHHHHHHHHHHHHHHHHHHHHH
❤❤❤
How can I know to switch the "a" term in the equation from foci and vertices without graphing
a > b, I think it's that simple
If the foci share the same x value (as with the 2nd comprehension example) the ellipse is vertical, so "a" has to be with y.
If the foci share the same y value (as with the 1st comprehension example) the ellipse is horizontal, so "a" has to be with x.
So if you want to know if the ellipse stretches vertical or horizontal all you need to know are the 2 foci, no need for vertices. The axis the foci share is the axis the ellipse stretches.
@@Teiwaz111I appreciate this clarification I find it more faster to determine whether the ellipse is horizontal or vertical using your instructions before I saw this I would first plot the center and count the units of x and y axis from left to right 😂
Does it matter if we express an equation of an ellipse as
(x^2)/(a^2) + (y^2)/(b^2) = 1
OR
(y^2)/(b^2) + (x^2)/(a^2) = 1 ? (since at the end of the day, both expressions are commutative)
Doesn't matter as - as you correctly say - both expressions are commutative.
It's just "standard notation" to start with the term that includes x.