Question:

Let x2a2+y2b2=1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 , where a>b a>b , be an ellipse whose eccentricity is 12 \frac{1}{\sqrt{2}} and the length of the latus rectum is 14 \sqrt{14} . Then the square of the eccentricity of x2a2y2b2=1 \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 is:

Updated On: Nov 15, 2024
  • 3
  • 72\frac{7}{2}
  • 32\frac{3}{2}
  • 52\frac{5}{2}
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The Correct Option is C

Solution and Explanation

We are given:

  • eccentricity e=12 e = \frac{1}{\sqrt{2}}
  • Length of the latus rectum L=14 L = \sqrt{14}

We know the formula for the latus rectum of an ellipse is:

L=2b2a L = \frac{2b^2}{a}

From the problem, we are given L=14 L = \sqrt{14} , so:

2b2a=14 \frac{2b^2}{a} = \sqrt{14}

Thus,

2b2=a14(1) 2b^2 = a \sqrt{14} \tag{1}

Next, use the relationship for the eccentricity e e of an ellipse:

e=1b2a2=12 e = \sqrt{1 - \frac{b^2}{a^2}} = \frac{1}{\sqrt{2}}

Squaring both sides:

e2=1b2a2=12 e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}

This implies:

b2a2=12 \frac{b^2}{a^2} = \frac{1}{2}

Substitute this into equation (1):

2(a22)=a14 2 \left(\frac{a^2}{2}\right) = a \sqrt{14} a2=a14 a^2 = a \sqrt{14}

Thus:

a=14 a = \sqrt{14}               1

Substitute a=14 a = \sqrt{14} into b2=a22 b^2 = \frac{a^2}{2} :

b2=142=7 b^2 = \frac{14}{2} = 7

Now, compute the square of the eccentricity:

e2=1b2a2=1714=714=12 e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{7}{14} = \frac{7}{14} = \frac{1}{2}

Thus, the square of the eccentricity is 32 \frac{3}{2} .

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