find a hyperbola equation with vertices (o,+-4) and passes through the point (3,-5)

Why did the hyperbola go see a psychiatrist?

Because it had a double focus!

Now, let's find the equation for your hyperbola. The standard form of a hyperbola equation with vertices at (0, ±4) is:

y^2/16 - x^2/a^2 = 1

To determine the value of "a" and complete the equation, we can use the fact that the hyperbola passes through the point (3, -5).

Plugging in the coordinates into the equation, we get:

(-5)^2/16 - 3^2/a^2 = 1
25/16 - 9/a^2 = 1

Multiplying through by 16a^2, we have:

25a^2 - 144 = 16a^2
9a^2 = 144
a^2 = 16

Therefore, the equation of the hyperbola is:

y^2/16 - x^2/16 = 1

To find the equation of a hyperbola with vertices and a point, we need to determine the values of the constants a, b, h, and k.

Given:
Vertices: (0, 4) and (0, -4)
Point on the hyperbola: (3, -5)

Step 1: Determine the center of the hyperbola.
The center of the hyperbola is the midpoint of the vertices. In this case, since the vertices have the same x-coordinate, the center is at (0, 0).

Therefore, the equation of the hyperbola is in the form: (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1

Step 2: Determine the value of a.
The value of a is the distance from the center to either of the vertices. In this case, a = 4.

Step 3: Determine the value of b.
We can find the value of b using the distance formula.
Given a vertex (0, 4) and the point on the hyperbola (3, -5), the distance can be calculated as:
√((x2 - x1)^2 + (y2 - y1)^2)

Using the values:
√((3 - 0)^2 + (-5 - 4)^2) = √(9 + 81) = √90

Since b is the distance from the center to the vertex, we need to halve √90 to find b:
b = √90 / 2 = √90 / √4 = √90 / 2√2 = √(90/4) = √(45/2) = √(45/2) * √2/√2 = (3√10) / 2

Step 4: Write the final equation of the hyperbola.
Plugging in the values into the equation, we get:
(x - 0)^2 / 4^2 - (y - 0)^2 / ((3√10) / 2)^2 = 1

Simplifying it further, we get:
x^2 / 16 - y^2 / (9(10/4)) = 1

This can be simplified to:
x^2 / 16 - y^2 / (45/2) = 1

So, the equation of the hyperbola is:
x^2 / 16 - y^2 / (45/2) = 1

To find the equation of a hyperbola, we need information about the vertices, a point on the hyperbola, and the distance between the foci. In this case, we have the vertices at (0, ±4) and a point on the hyperbola at (3, -5). We need to find the equation of the hyperbola.

First, let's determine the center of the hyperbola, which lies at the midpoint between the vertices:

Center = (0, (4 + (-4)) / 2) = (0, 0).

Now, let's find the distance between the center and one of the vertices, which gives us the value of a:

a = 4.

Next, we need to find the distance between the center and the given point (3, -5). This distance represents the value of c, and it can be calculated using the distance formula:

c = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(3 - 0)² + (-5 - 0)²]
= √[9 + 25]
= √34.

Now, we can determine the value of b using the relationship between a, b, and c:

c² = a² + b²
(√34)² = 4² + b²
34 = 16 + b²
b² = 34 - 16
b² = 18
b = √18 = 3√2.

Finally, we can write the equation of the hyperbola in standard form:

(x - h)² / a² - (y - k)² / b² = 1

Substituting the values, we get:

(x - 0)² / 4² - (y - 0)² / (3√2)² = 1

Simplifying,

x²/16 - y²/18 = 1

Therefore, the equation of the hyperbola passing through the point (3, -5) with vertices (0, ±4) is:

x²/16 - y²/18 = 1.

we know that the center is at (0,0) and the axis is vertical, so

y^2/a^2 - x^2/b^2 = 1
If the distance between the vertices is 2a = 8, then 4^2 + b^2 = c^2
Since (3,-5) is on the graph,

25/a^2 - 9/b^2 = 1
a=4, so
25/16 - 9/b^2 = 1
b^2 = 16

y^2/16 - x^2/16 = 1

See the graph at

http://www.wolframalpha.com/input/?i=plot+y%5E2%2F16+-+x%5E2%2F16+%3D+1