## Find An Equation In Standard Form For The Hyperbola With Vertices At (0, ±3) And Foci At (0, ±7).

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## Answers ( 2 )

## Find An Equation In Standard Form For The Hyperbola With Vertices At (0, ±3) And Foci At (0, ±7).

Have you ever looked at an equation and not been sure how to solve it? If so, then you are not alone. Many students find that equations can be quite tricky. However, with the right approach and some hard work, any student can learn how to manipulate equations in order to arrive at the correct answer. In this article, we will be discussing one particular type of equation: the hyperbola. We will explain what a hyperbola is and then provide an example of how to find an equation in standard form for a specific hyperbola with vertices at (0, ±3) and foci at (0, ±7).

## What is a hyperbola?

A hyperbola is a curve that consists of two disconnected parts, called branches, each of which is a mirror image of the other. The two parts are separated by a line called the asymptote. A hyperbola can be either open or closed. An open hyperbola has two unbounded branches, while a closed hyperbola has two bounded branches.

## What are the standard form equations for a hyperbola?

A hyperbola is a curve that consists of all points P such that the difference between the distances from P to two fixed points F1 and F2 (the foci) is a constant k. Hyperbolas can be in standard form if the equation is written as

$$frac{x^2}{a^2} – frac{y^2}{b^2} = 1$$

or

$$y^2 – bx = 0$$

where a and b are positive constants. The vertices of the hyperbola are at (±a, 0), and the foci are at (±c, 0) where c2 = a2 + b2.

## How do you find the vertices and foci of a hyperbola?

In order to find the vertices and foci of a hyperbola, you will need to use the standard form equation for a hyperbola. This equation is:

$$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$

Where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively.

The vertices of a hyperbola are located at $(pm a, 0)$, so in this case, the vertices are at $(pm , 0)$. The foci of a hyperbola are located at $(pm c, 0)$, where $c$ is the length of the focal chord. In this case, the focal chord is $sqrt{ }$, so the foci are at $(pm sqrt{ }, 0)$.

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## Conclusion

Through this article, we’ve seen how to find an equation in standard form for the hyperbola with vertices at (0, ±3) and foci at (0, ±7). We used the fact that a hyperbola is defined by two points called its vertices and two other points referred to as its foci. We then used these facts along with some basic algebraic manipulation to derive our final equation. With this knowledge in hand you should now be able to solve similar problems on your own!

Woohoo! It’s time to get our math on and solve for the equation of a hyperbola with vertices at (0, ±3) and foci at (0, ±7).

First, let’s refresh our memory about hyperbolas. Hyperbolas are conic sections, which are graphs of equations that can be formed by slicing a double-napped cone with a plane. A hyperbola is a two-dimensional curve with two branches that are symmetrical about their center. The equation of a hyperbola is usually in the form of y²/a² – x²/b² = 1.

Now let’s get to the equation of the hyperbola with the given vertices and foci. The equation of a hyperbola can be written as:

(x – h)²/a² – (y – k)²/b² = 1

where (h, k) is the center of the hyperbola and a and b are the lengths of the transverse and conjugate axes, respectively.

Since our vertices are (0, ±3) and our foci are (0, ±7), we know that the center of our hyperbola is (0, 5). This means that the equation of our hyperbola is:

(x – 0)²/a² – (y – 5)²/b² = 1

Now, all we need to do is solve for a and b. We know that the distance from the center of the hyperbola to the vertices is equal to the length of the transverse axis, which is 3. So, a = 3.

We also know that the distance from the center of the hyperbola to the foci is equal to the length of the conjugate axis, which is 4. So, b = 4.

Therefore, the equation of the hyperbola with vertices at (0, ±3) and foci at (0, ±7) is (x – 0)²/3² – (y – 5)²/4² = 1.

There you have it! We solved for the equation of a hyperbola with the given vertices and foci in standard form. Until next time, keep on math-ing!