Question 33 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: x²/100 + y²/400 = 1

Correct Answer

The given equation is x2100+y2400=1\frac{x^2}{100} + \frac{y^2}{400} = 1. Since the denominator of y2y^2 (400) is larger than the denominator of x2x^2 (100), the major axis is along the y-axis. Comparing with the standard form x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, we have a2=400a^2 = 400 and b2=100b^2 = 100, giving a=20a = 20 and b=10b = 10.

We calculate c=a2−b2=400−100=103c = \sqrt{a^2 - b^2} = \sqrt{400 - 100} = 10\sqrt{3}. Therefore, the coordinates of the foci are (0,±103)(0, \pm 10\sqrt{3}) and the vertices are (0,±20)(0, \pm 20). The length of the major axis is 2a=402a = 40 and the length of the minor axis is 2b=202b = 20.

The eccentricity is e=ca=10320=32e = \frac{c}{a} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}. The length of the latus rectum is 2b2a=20020=10\frac{2b^2}{a} = \frac{200}{20} = 10.

Exercise: EXERCISE 10.3 | Q: 6 | (Page 20)
For More Understanding

Explanation

The solution applies the standard properties of an ellipse where the major axis lies along the y-axis, as demonstrated in the provided context (specifically the example where the denominator of y2y^{2} is larger). The steps involve identifying 'a' and 'b' from the denominators, calculating 'c' for the foci, and determining eccentricity and latus rectum using the standard formulas derived from the text.

Solution Steps

  1. Step 1: Identify the major axis. Since the denominator of y2y^2 (400) is larger than the denominator of x2x^2 (100), the major axis is along the y-axis.

  2. Step 2: Find a and b. Comparing x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, we get a2=400  ⟹  a=20a^2 = 400 \implies a = 20 and b2=100  ⟹  b=10b^2 = 100 \implies b = 10.

  3. Step 3: Calculate c. c=a2−b2=400−100=300=103c = \sqrt{a^2 - b^2} = \sqrt{400 - 100} = \sqrt{300} = 10\sqrt{3}.

  4. Step 4: Determine coordinates. Foci are (0,±c)=(0,±103)(0, \pm c) = (0, \pm 10\sqrt{3}) and Vertices are (0,±a)=(0,±20)(0, \pm a) = (0, \pm 20).

  5. Step 5: Calculate lengths and eccentricity. Major axis = 2a=402a = 40, Minor axis = 2b=202b = 20, Eccentricity e=ca=32e = \frac{c}{a} = \frac{\sqrt{3}}{2}, Latus rectum = 2b2a=10\frac{2b^2}{a} = 10.

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