Question 39 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: 9y² - 4x² = 36

Correct Answer

The given equation is 9y2−4x2=369y^2 - 4x^2 = 36.

Step 1: Convert to standard form Dividing both sides by 36: y24−x29=1\frac{y^2}{4} - \frac{x^2}{9} = 1 This is of the form y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1, so the transverse axis is along y-axis.

Step 2: Identify a and b a2=4a^2 = 4, so a=2a = 2 b2=9b^2 = 9, so b=3b = 3

Step 3: Find c and eccentricity c2=a2+b2=4+9=13c^2 = a^2 + b^2 = 4 + 9 = 13, so c=13c = \sqrt{13} Eccentricity e=ca=132e = \frac{c}{a} = \frac{\sqrt{13}}{2}

Step 4: Find vertices and foci Vertices: (0,±a)=(0,±2)(0, \pm a) = (0, \pm 2) Foci: (0,±c)=(0,±13)(0, \pm c) = (0, \pm \sqrt{13})

Step 5: Find latus rectum Length of latus rectum =2b2a=2×92=9= \frac{2b^2}{a} = \frac{2 \times 9}{2} = 9

Answer: Vertices: (0,±2)(0, \pm 2); Foci: (0,±13)(0, \pm \sqrt{13}); Eccentricity: 132\frac{\sqrt{13}}{2}; Latus rectum: 99

Exercise: EXERCISE 10.4 | Q: 3 | (Chapter: 27)
For More Understanding

Explanation

The question asks to find properties of hyperbola 9y² - 4x² = 36. Following the textbook method shown in Example 14 and Example 15, we first convert to standard form. Since the positive term has y², the transverse axis is along y-axis (as stated in the context: 'It is the positive term whose denominator gives the transverse axis'). Using the standard formulas: vertices at (0, ±a), foci at (0, ±c) where c² = a² + b², eccentricity e = c/a, and latus rectum = 2b²/a.

Solution Steps

  1. Step 1: Convert equation to standard form by dividing by 36

  2. Step 2: Identify a² = 4, b² = 9, so a = 2, b = 3

  3. Step 3: Calculate c² = a² + b² = 13, so c = 13\sqrt{13}

  4. Step 4: Find vertices (0, ±2) and foci (0, ±13\sqrt{13})

  5. Step 5: Calculate eccentricity e = 13\sqrt{13}/2 and latus rectum = 9

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