Question 35 of 70intermediate🔧 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: 16x² + y² = 16

Correct Answer

The given equation is 16x² + y² = 16. Dividing both sides by 16, we get:

x21+y216=1\frac{x^2}{1} + \frac{y^2}{16} = 1

Comparing with the standard equation x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, we have b² = 1 and a² = 16, so b = 1 and a = 4.

Since the denominator of y216\frac{y^2}{16} (which is 16) is larger than the denominator of x21\frac{x^2}{1} (which is 1), the major axis is along the y-axis.

Now, c = a2−b2\sqrt{a² - b²} = 16−1\sqrt{16 - 1} = 15\sqrt{15}

Eccentricity e = c/a = 15\sqrt{15}/4

Foci: (0, ±c) = (0, ±15\sqrt{15})

Vertices: (0, ±a) = (0, ±4)

Length of major axis = 2a = 8 units

Length of minor axis = 2b = 2 units

Length of latus rectum = 2b²/a = 2(1)/4 = 1/2 units

Exercise: EXERCISE 10.3 | Q: 8 | (Page 20)
For More Understanding

Explanation

This question requires converting the ellipse equation to standard form by dividing by 16. The key insight is recognizing that since the denominator under y² (16) is larger than the denominator under x² (1), the major axis lies along the y-axis. This means we use the form x²/b² + y²/a² = 1 where a > b. The solution follows the same pattern as Example 9 in the textbook context.

Solution Steps

  1. Step 1: Convert equation to standard form by dividing by 16: x²/1 + y²/16 = 1

  2. Step 2: Identify a² = 16, b² = 1, so a = 4, b = 1

  3. Step 3: Since denominator under y² is larger, major axis is along y-axis

  4. Step 4: Calculate c = a2−b2\sqrt{a² - b²} = 16−1\sqrt{16-1} = 15\sqrt{15}

  5. Step 5: Calculate eccentricity e = c/a = 15\sqrt{15}/4

  6. Step 6: Find foci (0, ±15\sqrt{15}), vertices (0, ±4)

  7. Step 7: Calculate major axis = 8 units, minor axis = 2 units, latus rectum = 1/2 units

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