Question 56 of 70beginner🔧 ApplyShort Answer2 marks

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions: Vertices (0, ± 3), foci (0, ± 5)

Correct Answer

Since the foci are on the y-axis, the equation of the hyperbola is of the form

y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

Given vertices are (0,±3)(0, \pm 3), so a=3a = 3.

Given foci are (0,±5)(0, \pm 5), so c=5c = 5.

Using the relation b2=c2−a2b^2 = c^2 - a^2:

b2=25−9=16b^2 = 25 - 9 = 16

Therefore, the equation of the hyperbola is:

y29−x216=1\frac{y^2}{9} - \frac{x^2}{16} = 1
Exercise: EXERCISE 10.4 | Q: 9 | (Page 27)
For More Understanding

Explanation

This question is from Exercise 10.4, Question 9. The solution follows the same pattern as Example 15 in the textbook. Since both vertices and foci have coordinates of the form (0, ±k), the transverse axis lies on the y-axis, which means the hyperbola equation takes the form y²/a² - x²/b² = 1. The key relationship for hyperbolas is b² = c² - a², which differs from the ellipse relationship.

Solution Steps

  1. Step 1: Identify that foci on y-axis means equation form is y²/a² - x²/b² = 1

  2. Step 2: From vertices (0, ±3), determine a = 3

  3. Step 3: From foci (0, ±5), determine c = 5

  4. Step 4: Calculate b² = c² - a² = 25 - 9 = 16

  5. Step 5: Write the equation: y²/9 - x²/16 = 1

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