Question 52 of 70intermediate🔧 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: 5y² - 9x² = 36

Correct Answer

The given equation is 5y² - 9x² = 36. Dividing both sides by 36, we get: y²/(36/5) - x²/4 = 1

This is of the form y²/a² - x²/b² = 1, where a² = 36/5 and b² = 4. So a = 6/5\sqrt{5} and b = 2.

Vertices: Since the transverse axis is along y-axis, vertices are (0, ±a) = (0, ±6/5\sqrt{5}).

Foci: c² = a² + b² = 36/5 + 4 = 56/5, so c = 56/5\sqrt{56/5} Foci are (0, ±c) = (0, ±56/5\sqrt{56/5}).

Eccentricity: e = c/a = 56/5\sqrt{56/5} ÷ (6/5\sqrt{5}) = 56\sqrt{56}/6 = 14\sqrt{14}/3

Length of latus rectum: 2b²/a = 2(4)/(6/5\sqrt{5}) = 45\sqrt{5}/3

Exercise: EXERCISE 10.4 | Q: 5 | (Page 27)
For More Understanding

Explanation

This question requires converting the hyperbola equation to standard form. Since the y² term is positive, the transverse axis is along the y-axis. The standard form y²/a² - x²/b² = 1 is used. Key formulas from the context: c² = a² + b² for hyperbola, eccentricity e = c/a, and length of latus rectum = 2b²/a.

Solution Steps

  1. Step 1: Divide equation by 36 to get y²/(36/5) - x²/4 = 1

  2. Step 2: Identify a² = 36/5, b² = 4, so a = 6/5\sqrt{5}, b = 2

  3. Step 3: Calculate c² = a² + b² = 56/5, so c = 56/5\sqrt{56/5}

  4. Step 4: Vertices are (0, ±a) = (0, ±6/5\sqrt{5})

  5. Step 5: Foci are (0, ±c) = (0, ±56/5\sqrt{56/5})

  6. Step 6: Eccentricity e = c/a = 14\sqrt{14}/3

  7. Step 7: Length of latus rectum = 2b²/a = 45\sqrt{5}/3

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