Endpoints of Conjugate axis (3,-2), (3,6), length of transverse units is 16 units

To find the equation of the hyperbola, you are given the following information:

- The endpoints of the conjugate axis are \((3, -2)\) and \((3, 6)\).
- The length of the transverse axis is 16 units.

### Step 1: Find the center of the hyperbola

The center of the hyperbola is the midpoint of the conjugate axis. The midpoint between \((3, -2)\) and \((3, 6)\) can be calculated as:

\[
\text{Midpoint} = \left( \frac{3+3}{2}, \frac{-2+6}{2} \right) = (3, 2)
\]

So, the center of the hyperbola is \((3, 2)\).

### Step 2: Determine the distance of the conjugate axis

The distance between the endpoints of the conjugate axis is the length of the conjugate axis. The distance between \((3, -2)\) and \((3, 6)\) is:

\[
\text{Distance} = \sqrt{(3 - 3)^2 + (6 - (-2))^2} = \sqrt{0^2 + (6 + 2)^2} = \sqrt{64} = 8
\]

Thus, the length of the conjugate axis is 8 units. Since the length of the conjugate axis is \(2b\), we have:

\[
2b = 8 \quad \Rightarrow \quad b = 4
\]

### Step 3: Determine the value of \(a\)

The length of the transverse axis is given as 16 units. The length of the transverse axis is \(2a\), so:

\[
2a = 16 \quad \Rightarrow \quad a = 8
\]

### Step 4: Write the equation of the hyperbola

Since the conjugate axis is vertical (aligned with the y-axis), and the transverse axis is horizontal (aligned with the x-axis), the standard form of the equation of the hyperbola is:

\[
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]

where \((h, k)\) is the center of the hyperbola, \(a\) is half the length of the transverse axis, and \(b\) is half the length of the conjugate axis. Substituting the known values:

- \(h = 3\), \(k = 2\),
- \(a = 8\), so \(a^2 = 64\),
- \(b = 4\), so \(b^2 = 16\).

Thus, the equation of the hyperbola is:

\[
\frac{(x - 3)^2}{64} - \frac{(y - 2)^2}{16} = 1
\]

### Final Answer:
The equation of the hyperbola is:

\[
\frac{(x - 3)^2}{64} - \frac{(y - 2)^2}{16} = 1
\]

Would you like further clarification or more details?

### Related Questions:
1. How do you find the foci of this hyperbola?
2. What is the eccentricity of the hyperbola?
3. How do you sketch the hyperbola based on this equation?
4. How is the equation of a vertical hyperbola different from a horizontal one?
5. What is the general relationship between the transverse and conjugate axes in a hyperbola?

### Tip:
Remember that for hyperbolas, the relationship \(c^2 = a^2 + b^2\) helps to find the distance to the foci, where \(c\) is the distance from the center to each focus.

Learn how to find the equation of a hyperbola with conjugate axis endpoints (3,-2) and (3,6), and a transverse axis length of 16 units. This step-by-step guide covers the midpoint formula, distance formula, and the hyperbola equation. Ideal for students in Grades 10-12.

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