Question 55 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, ± 5), foci (0, ± 8)

Correct Answer

The equation of the hyperbola is y225−x239=1\frac{y^2}{25} - \frac{x^2}{39} = 1.

Exercise: EXERCISE 10.4 | Q: 8 | (Page 27)
For More Understanding

Explanation

This question is from Exercise 10.4, Question 8. Since the vertices (0, ±5) and foci (0, ±8) both lie on the y-axis, the hyperbola has its transverse axis along the y-axis. The standard form follows the pattern shown in Example 15 of the context, where the equation is y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1. Using the relationship c2=a2+b2c^2 = a^2 + b^2, we can find b² and write the complete equation.

Solution Steps

  1. Step 1: Since vertices are (0, ±5) and foci are (0, ±8), both lie on the y-axis. Therefore, the hyperbola has the form y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1.

  2. Step 2: From vertices (0, ±5), we get a = 5, so a² = 25.

  3. Step 3: From foci (0, ±8), we get c = 8, so c² = 64.

  4. Step 4: Using the relation c2=a2+b2c^2 = a^2 + b^2, we get b2=c2−a2=64−25=39b^2 = c^2 - a^2 = 64 - 25 = 39.

  5. Step 5: Substituting a² = 25 and b² = 39, the equation of the hyperbola is y225−x239=1\frac{y^2}{25} - \frac{x^2}{39} = 1.

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