Question 49 of 70beginner🔧 ApplyShort Answer2 marks

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: y²/9 - x²/27 = 1

Correct Answer

The given equation is y²/9 - x²/27 = 1, which is of the form y²/a² - x²/b² = 1. Here, a² = 9 and b² = 27, so a = 3. For a hyperbola with transverse axis along y-axis: c² = a² + b² = 9 + 27 = 36, so c = 6.

Vertices: (0, ±a) = (0, ±3)

Foci: (0, ±c) = (0, ±6)

Eccentricity: e = c/a = 6/3 = 2

Length of latus rectum: 2b²/a = 2(27)/3 = 18

Exercise: EXERCISE 10.4 | Q: 2 | (Page 27)
For More Understanding

Explanation

This question is from Exercise 10.4, Question 2. The context shows that for a hyperbola in the form y²/a² - x²/b² = 1, the transverse axis is along the y-axis (as stated: 'It is the positive term whose denominator gives the transverse axis'). The vertices are at (0, ±a), foci at (0, ±c) where c² = a² + b², and the length of latus rectum is 2b²/a. Example 15 in the context demonstrates solving similar hyperbola problems with foci on the y-axis.

Solution Steps

  1. Step 1: Compare y²/9 - x²/27 = 1 with standard form y²/a² - x²/b² = 1 to get a² = 9, b² = 27, a = 3

  2. Step 2: Calculate c² = a² + b² = 9 + 27 = 36, so c = 6

  3. Step 3: Vertices are (0, ±a) = (0, ±3)

  4. Step 4: Foci are (0, ±c) = (0, ±6)

  5. Step 5: Eccentricity e = c/a = 6/3 = 2

  6. Step 6: Length of latus rectum = 2b²/a = 2(27)/3 = 18

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