The length of the latus rectum for a hyperbola is given by the formula (2b²)/(a). We are given that this length is 36. Setting up the equation, we get (2b²)/(a) = 36. Simplifying this equation, we find b² = 18a.

The length of the latus rectum for a hyperbola is given by the formula (2b²)/(a). We are given that this length is 36. Setting up the equation, we get (2b²)/(a) = 36. Simplifying this equation, we find b² = 18a.

For a hyperbola, the relationship between a, b, and c is c² = a² + b². We know c = 12 and we found b² = 18a. Substituting these values into the equation, we get 12² = a² + 18a, which simplifies to 144 = a² + 18a. Rearranging this into a standard quadratic form gives a² + 18a - 144 = 0.

For a hyperbola, the relationship between a, b, and c is c² = a² + b². We know c = 12 and we found b² = 18a. Substituting these values into the equation, we get 12² = a² + 18a, which simplifies to 144 = a² + 18a. Rearranging this into a standard quadratic form gives a² + 18a - 144 = 0.

Find the equation of the hyperbola where foci are $(0, \pm 12)$ and the length of the latus rectum is 36

Answer: The equation of the hyperbola is $\frac{y^2}{36} - \frac{x^2}{108} = 1$ .

Step-by-step solution

Step 1: Identify Foci and Standard Form

The given foci are $(0, \pm 12)$ . For a hyperbola, the foci are at $(0, \pm c)$ for a vertical hyperbola, or $(\pm c, 0)$ for a horizontal hyperbola. Since the $x$ -coordinate is 0, the foci are on the $y$ -axis, indicating a vertical hyperbola. Therefore, $c = 12$ . The standard equation for a vertical hyperbola centered at the origin is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ .

Step 2: Use Latus Rectum Length

The length of the latus rectum for a hyperbola is given by the formula $\frac{2b^2}{a}$ . We are given that this length is 36. Setting up the equation, we get $\frac{2b^2}{a} = 36$ . Simplifying this equation, we find $b^2 = 18a$ .

Step 3: Relate a, b, and c

For a hyperbola, the relationship between $a$ , $b$ , and $c$ is $c^2 = a^2 + b^2$ . We know $c = 12$ and we found $b^2 = 18a$ . Substituting these values into the equation, we get $12^2 = a^2 + 18a$ , which simplifies to $144 = a^2 + 18a$ . Rearranging this into a standard quadratic form gives $a^2 + 18a - 144 = 0$ .

Step 4: Solve for 'a'

We solve the quadratic equation $a^2 + 18a - 144 = 0$ for $a$ . Factoring the quadratic, we find $(a + 24)(a - 6) = 0$ . This gives two possible values for $a$ : $a = -24$ or $a = 6$ . Since $a$ represents a distance, it must be positive, so we choose $a = 6$ .

Step 5: Calculate 'b²'

Now that we have $a = 6$ , we can find $b^2$ using the relationship $b^2 = 18a$ that we derived from the latus rectum length. Substituting $a = 6$ into this equation, we get $b^2 = 18(6)$ , which simplifies to $b^2 = 108$ .

Step 6: Write the Equation of the Hyperbola

We have determined that $a = 6$ (so $a^2 = 36$ ) and $b^2 = 108$ . Substitute these values into the standard equation for a vertical hyperbola: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ . This gives us the final equation of the hyperbola: $\frac{y^2}{36} - \frac{x^2}{108} = 1$ .

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Find the equation of the hyperbola where foci are (0,±12)(0, \pm 12)(0,±12) and the length of the latus rectum is 36