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Precalculus
Vertices & direction of a hyperbola (example 2)
Sal matches an equation to a given graph of a hyperbola, based on the hyperbola's direction & vertices.
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How exactly would you find the a^2 under the y^2 other than just testing points? Is it possible to algebraically find it?(19 votes)
Bit late, but for anyone wondering, you cannot. You have two variables, y and a, therefore you need two equations to solve it. We only have one so there are infinite solutions.(1 vote)
What value does the nine ( 3 squared ) hold in the graph? What is it's use?(4 votes)
Since this hyperbola opens left and right, the number 3^2 (=b^2 in the standard formula) doesn't designate a point, the way a=5 does. It represents the rise and fall of the asymptotes with respect to a run of 5: the asymptotes have slopes of +/- 3/5. As for it use: I would use it to draw the asymptotes, which make it easy (or easier) to draw the curve correctly.(7 votes)
Since it's been like this in all the videos, I just have a question. Are all hyperbolas always centered around the origin?(3 votes)
Not at all, we can shift them vertically and horizontally like any graph. Just replace y with y+c to shift down by c, and replace x with x+c to shift left by c.(5 votes)
How to solve the asymptote line of the hyperbole(3 votes)
If you know the values of a and b and you know the orientation of the hyperbola, or if you know the equation of the hyperbola, the asymptote has a slope based on a and b. If you do not know a and b, directly, but only the coordinates of the vertices and foci, then you have to solve for the missing lengths. The distance from the center to each vertex is a. The distance from the center to each focus is c. You can obtain the length of b by using Pythagoras, c² = a² + b², so that b = √(c² - a²)
Let's start with a hyperbola with a center at the origin (0,0)
A hyperbola that opens to the sides (transverse axis is horizontal, the x-axis) has an equation
x²/a² - y²/b² = 1
Then, the asymptotes are the lines:
y = b/a x and y = - b/a x
A hyperbola that opens up and down (transverse axis is vertical, the y-axis) has the equation
y²/a² - x²/b² = 1
Then, the asymptotes are the lines:
y = a/b x and y = - a/b x
If the hyperbola is shifted (but not tilted), then the equations are more complicated:
A hyperbola that opens to the sides (transverse axis is horizontal, parallel to the x-axis) with a center at (h, k) has an equation
(x - h)²/a² - (y - k)²/b² = 1
Then, the asymptotes are the lines:
y - k = ±b/a (x - h)
A hyperbola is shifted (but not tilted), and that opens up and down (transverse axis is vertical, parallel to the y-axis) ) with a center at (h, k) has the equation
(y - k)²/a² - (x - h)²/b² = 1
Then, the asymptotes are the lines:
y - k = ±a/b (x - h)(3 votes)
- @1:35I don't understand why Sal is writing a and 25 under two separate fractions when a is 5. Or why a is placed under y, that's different from standard form that I'm being taught right now.(2 votes)
Could the value of X be negative even if it still came in that same form (X^2/25 - Y^2/9)?(2 votes)- Good question! You can see that x can be negative: the entire left wing of the curve is in quadrants 2 and 3, where x is negative.(1 vote)
At1:08, why does he call it a^2? I thought that a was the center to either vertex, not the other one. Please and thank you!!(1 vote)- Sal is only assigning a random variable for that denominator.(2 votes)
hi, at0:51, since when it has been reasoned that the distance from the vertices to the centre is 5, then the term under x^2 in the equation is 5^2? Cheers.(1 vote)- Yes, that's correct. At0:51in the segment, the speaker reasoned that the distance from the vertices to the center of the hyperbola is 5 units in the horizontal direction. Since the standard form of the equation of a hyperbola is ((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1 for a hyperbola centered at (h, k), and the hyperbola is centered at (0,0), the value of a^2 (which represents the distance from the center to the vertices in the horizontal direction) can be found by squaring the distance, which in this case is 5. Therefore, the term under x^2 in the equation is 5^2 or 25.(2 votes)
- graph seems abstract, what is it used for in real life applications?(1 vote)
- Hyperbolas have many real-life applications in various fields such as engineering, physics, optics, and finance. Here are some examples:
Communication systems: Hyperbolas can be used to determine the location of a source of radio signals using two or more antennas. This technique is used in GPS (Global Positioning System) to determine the location of a receiver on Earth.
Optics: Hyperbolas are used to model the shape of some mirrors and lenses, such as the parabolic mirrors used in telescopes and satellite dishes.
Finance: Hyperbolas can be used to model the relationship between two variables, such as the price and demand of a product. The resulting hyperbola is called a hyperbolic curve and is used in finance to model the yield curve of bonds.
Projectile motion: The path of a projectile launched at an angle to the horizontal is a hyperbola if the air resistance is negligible.
Nuclear physics: Hyperbolas are used to model the relationship between energy and momentum in nuclear reactions.
Overall, hyperbolas have a wide range of applications in different fields of science, engineering, and mathematics.(1 vote)
can the distance of the vertices be different?(1 vote)- no, they always have to be the same
if I understand your question correctly(1 vote)
1:35Video transcript
- [Voiceover] So we're asked to choose the equation that can represent the hyperbola graphed below. And so this is the hyperbola graphed in blue and I encourage
you to pause the video and figure out which of these equations are represented by the graph here. Alright, so let's think about it. This graph opens to
the left and the right. Well, I guess, the first
thing we could realize, it's centered at zero. So definitely, it's just going to have the form, x squared and y squared over two different things equaling one. And we know that it opens
to the left and the right. You can think of it opens
along the x-direction and so we know that the x-term is going to be positive here,
which tells us that the y-term is going to be negative. And we know that the vertices here are five to the right of the center and five to the left of the center and so since the distance
from the vertices to the center is five in
the horizontal direction, we know that this right over here is going to be five squared or 25. And this we don't quite know, just yet. I just call this a squared.
We don't know what a is. Now let's look at these choices here. X squared over 25 minus y squared over nine equals one. Well, that seems to match the pattern that I was able to
generate really quickly, just looking at the
graph, so I like this one. This has the x-term being negative. So this graph over
here, this would open up up and down, not to
the left and the right, so we could rule this out. This one over here. This
has x squared over nine. That would imply that our x-intercepts are plus or minus three
to the right and left of the center, not five. Clearly, they aren't plus or minus three so we could rule this one out. This one has the y-term being positive and the x-term being negative, so once again, this
would open up and down. So we could rule that one out as well. And so our first choice
that we like that matched our pattern, we could
feel pretty good about it. Now if you wanted to verify the nine or if you want, you might want to try out some other points or solve some points, if it wasn't multiple
choice but in this case, we are able to pick out. This is the only one that even matches the general structure that
we were able to deduce.