Question 28 of 70beginner🔧 ApplyShort Answer2 marks

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse: x²/36 + y²/16 = 1

Correct Answer

Given equation of ellipse: x²/36 + y²/16 = 1

Comparing with standard form x²/a² + y²/b² = 1, since denominator of x² (36) is larger than denominator of y² (16), the major axis is along x-axis.

a² = 36, so a = 6 b² = 16, so b = 4

c = a2−b2\sqrt{a² - b²} = 36−16\sqrt{36 - 16} = 20\sqrt{20} = 25\sqrt{5}

Eccentricity (e) = c/a = 25\sqrt{5}/6 = 5\sqrt{5}/3

Coordinates of foci: (±25\sqrt{5}, 0) i.e., (25\sqrt{5}, 0) and (-25\sqrt{5}, 0)

Vertices: (±6, 0) i.e., (6, 0) and (-6, 0)

Length of major axis = 2a = 12 units

Length of minor axis = 2b = 8 units

Length of latus rectum = 2b²/a = 2(16)/6 = 16/3 units

Exercise: EXERCISE 10.3 | Q: 1 | (Page 20)
For More Understanding

Explanation

This is Exercise 10.3, Question 1 from the NCERT textbook. The solution follows the standard method shown in Example 9 of the same chapter. When the denominator of x² is larger, the major axis lies along the x-axis, and we use the standard form x²/a² + y²/b² = 1. The key formulas used are: c = a2−b2\sqrt{a² - b²} for finding the distance of foci from center, e = c/a for eccentricity, and length of latus rectum = 2b²/a.

Solution Steps

  1. Step 1: Identify a² = 36, b² = 16 by comparing with standard form x²/a² + y²/b² = 1

  2. Step 2: Calculate a = 6, b = 4

  3. Step 3: Find c = a2−b2\sqrt{a² - b²} = 20\sqrt{20} = 25\sqrt{5}

  4. Step 4: Calculate eccentricity e = c/a = 5\sqrt{5}/3

  5. Step 5: Foci are (±c, 0) = (±25\sqrt{5}, 0)

  6. Step 6: Vertices are (±a, 0) = (±6, 0)

  7. Step 7: Major axis = 2a = 12, Minor axis = 2b = 8

  8. Step 8: Latus rectum = 2b²/a = 16/3

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