Question 40 of 70beginner🔧 ApplyShort Answer2 marks

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions: Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)

Correct Answer

Since the ends of the major axis are (±3,0)(\pm 3, 0), the major axis lies along the x-axis. Therefore, the equation of the ellipse is of the standard form x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

From the given coordinates, the ends of the major axis correspond to (±a,0)(\pm a, 0), which implies a=3a = 3. Similarly, the ends of the minor axis are (0,±2)(0, \pm 2), corresponding to (0,±b)(0, \pm b), which implies b=2b = 2.

Substituting the values a2=9a^2 = 9 and b2=4b^2 = 4 into the standard equation, the required equation is x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1.

Exercise: EXERCISE 10.3 | Q: 13 | (Page 20)
For More Understanding

Explanation

The solution follows the logic presented in Example 11 of the provided context. Since the ends of the major axis are on the x-axis (y=0), the major axis is horizontal, dictating the standard equation form where 'a' is associated with the x-term. The values for 'a' and 'b' are derived directly from the coordinates of the axis ends provided in the question.

Solution Steps

  1. Step 1: Identify the orientation of the axes. Since the ends of the major axis are (±3,0)(\pm 3, 0), the major axis lies along the x-axis.

  2. Step 2: Determine the values of 'a' and 'b'. The ends of the major axis are (±a,0)(\pm a, 0), so a=3a = 3. The ends of the minor axis are (0,±b)(0, \pm b), so b=2b = 2.

  3. Step 3: Write the standard equation. Since the major axis is along the x-axis, the equation is x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

  4. Step 4: Substitute the values. Substituting a2=9a^2 = 9 and b2=4b^2 = 4, the equation becomes x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1.

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