Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) The eccentricity of the ellipse \[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \] is \( \dfrac{3}{__} \).

Hint:

For an ellipse with equation \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) where \(a>b\), eccentricity is always \[ e=\sqrt{1-\frac{b^2}{a^2}}. \]

Answer 1

Correct Answer: 5

Step 1: Compare the given equation with the standard form of an ellipse: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Here, \[ a^2 = 25 \Rightarrow a = 5,\quad b^2 = 16 \Rightarrow b = 4 \]
Step 2: Eccentricity \( e \) of an ellipse is given by: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Step 3: Substitute the values: \[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]
Step 4: Comparing with the given form \( \dfrac{3}{__} \), the denominator is: \[ 5 \] Final Answer (up to two decimal places): \[ \boxed{5.00} \]