Question

standard form of the equation of a hyperbola that has vertices at (-10, -15) and (70, -15) and one of its foci at (-11, -15).

Answer 1

The standard form of the equation of the hyperbola with vertices at (-10, -15) and (70, -15) and one of its foci at (-11, -15) is (x - 30)^2 / 1600 - (y + 15)^2 / 81 = 1

Step-by-step explanation:

To write the standard form of the equation of a hyperbola with the given vertices and a focus, we'll follow these steps:

1. Determine the center of the hyperbola.
2. Calculate the distance between the vertices and the center to find the length of the transverse axis (2a).
3. Calculate the distance between a focus and the center to find the focal distance (c).
4. Use the relationship c^2 = a^2 + b^2 to determine the length of the conjugate axis (2b).
5. Write the standard form equation based on the orientation of the hyperbola.

Step 1: Determine the center of the hyperbola.
The center of the hyperbola is the midpoint of the line segment joining the two vertices. Since the vertices are at (-10, -15) and (70, -15), the center (h, k) can be found as follows:

h = (-10 + 70) / 2 = 60 / 2 = 30
k = (-15 + (-15)) / 2 = -30 / 2 = -15

So, the center of the hyperbola is at (30, -15).

Step 2: Calculate the length of the transverse axis (2a).
The distance between the vertices is the length of the transverse axis. The vertices are 80 units apart because they are at (-10, -15) and (70, -15). This means:

2a = 80
a = 40

Therefore, the length of the semi-transverse axis a is 40 units.

Step 3: Calculate the focal distance (c).
The focal distance is the distance between the center and one of the foci. We were given one focus at (-11, -15). Since the center is at (30, -15), the focal distance c is:

c = |30 - (-11)| = |30 + 11| = 41

Step 4: Use the relationship c^2 = a^2 + b^2 to determine b.
We know that a = 40 and c = 41. Plugging these values into the relationship gives us:

41^2 = 40^2 + b^2
1681 = 1600 + b^2
b^2 = 1681 - 1600
b^2 = 81
b = 9

Therefore, the length of the semi-conjugate axis b is 9 units.

Step 5: Write the standard form equation.
Since the hyperbola is horizontal (the vertices have the same y-coordinate), the standard form of its equation is:

(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1

Plugging in the values for h, k, a, and b, we get:

(x - 30)^2 / 40^2 - (y + 15)^2 / 9^2 = 1

Simplify further by squaring the values of a and b:

(x - 30)^2 / 1600 - (y + 15)^2 / 81 = 1

This is the standard form of the equation of the hyperbola with vertices at (-10, -15) and (70, -15) and one of its foci at (-11, -15).

Answer 2

check the picture below. So with those points, ther hyperbola looks more or less like so.

now, notice the distance from vertex to vertex, half-way is the center, at 30, -15.

now, from the center to the given focus point, at -11, - 15, is 41 units, bear in mind the distance from the center to either foci is "c". Thus c = 41

since the traverse axis for the hyperbola is the x-axis, thus the fraction with the "x" variable is the positive fraction. thus


`\bf \textit{hyperbolas, horizontal traverse axis }\\\\ \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1 \qquad \begin{cases} center\ ({{ h}},{{ k}})\\ vertices\ ({{ h}}\pm a, {{ k}})\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{{{ a }}^2+{{ b }}^2} \end{cases}`


`\bf \begin{cases} h=30\\ k=-15\\ a=40 \end{cases}\implies \cfrac{(x-30)^2}{30^2}-\cfrac{[y-(-15)]^2}{b^2}=1 \\\\\\ \cfrac{(x-30)^2}{30^2}-\cfrac{(y+15)^2}{b^2}=1\\\\ -------------------------------\\\\ \textit{now, we know }c=41\implies c=√(a^2+b^2)\implies c^2=a^2+b^2 \\\\\\ √(c^2-a^2)=b\implies √(41^2-30^2)=b\implies √(781)=b\\\\ -------------------------------\\\\ \cfrac{(x-30)^2}{30^2}-\cfrac{(y+15)^2}{(√(781))^2}=1\implies \cfrac{(x-30)^2}{900}-\cfrac{(y+15)^2}{781}=1`