Question 58 of 70beginner🔧 ApplyShort Answer2 marks

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions: Foci (0, ± 13), the conjugate axis is of length 24.

Correct Answer

The equation of the hyperbola is y225−x2144=1\frac{y^2}{25} - \frac{x^2}{144} = 1.

Exercise: EXERCISE 10.4 | Q: 11 | (Page 27)
For More Understanding

Explanation

This question is from Exercise 10.4, Question 11. Since the foci are at (0, ±13), the hyperbola has its transverse axis on the y-axis, so the equation takes the form y²/a² - x²/b² = 1. The conjugate axis length gives us 2b = 24, so b = 12 and b² = 144. Using the relation c² = a² + b² for hyperbolas, we find a² = 25.

Solution Steps

  1. Step 1: Since foci are (0, ±13), the hyperbola has transverse axis on y-axis. Equation form: y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

  2. Step 2: From foci, c = 13, so c² = 169.

  3. Step 3: Conjugate axis length = 2b = 24, therefore b = 12 and b² = 144.

  4. Step 4: Using relation c² = a² + b²: 169 = a² + 144, therefore a² = 25.

  5. Step 5: Substituting values: y225−x2144=1\frac{y^2}{25} - \frac{x^2}{144} = 1

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