Question

If (-1, 0) and (3, 0) are foci of an ellipse and the length of the major axis is 6, then the length of the minor axis is

• A. √5
• B. 5
• C. 10
• D. 2√5
• E. 3

Answer 1

The Correct Option is D

The equation of an ellipse with foci at \( (c, 0) \) and \( (-c, 0) \) and semi-major axis \( a \) and semi-minor axis \( b \) is given by: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where the relationship between the semi-major axis \( a \), semi-minor axis \( b \), and the distance of the foci \( c \) is: \[ c^2 = a^2 - b^2 \] We are given the foci at \( (-1, 0) \) and \( (3, 0) \), so the distance between the foci is: \[ 2c = 3 - (-1) = 4 \quad \Rightarrow \quad c = 2 \] We are also given that the length of the major axis is 6, so: \[ 2a = 6 \quad \Rightarrow \quad a = 3 \] Now, using the equation \( c^2 = a^2 - b^2 \), we can solve for \( b^2 \): \[ 2^2 = 3^2 - b^2 \quad \Rightarrow \quad 4 = 9 - b^2 \quad \Rightarrow \quad b^2 = 5 \] Thus, the length of the minor axis is: \[ 2b = 2\sqrt{5} \]

The correct option is (D) : \(2\sqrt 5\)

Answer 2

The foci of the ellipse are (-1, 0) and (3, 0). The center of the ellipse is the midpoint of the segment connecting the foci. The midpoint is \(\left(\frac{-1+3}{2}, \frac{0+0}{2}\right) = (1, 0)\).

The distance between the foci is \(2c\), where \(c\) is the distance from the center to each focus. The distance between (-1, 0) and (3, 0) is \(\sqrt{(3 - (-1))^2 + (0 - 0)^2} = \sqrt{4^2} = 4\). So, \(2c = 4\), which means \(c = 2\).

The length of the major axis is \(2a\), where \(a\) is the length of the semi-major axis. We are given that the length of the major axis is 6, so \(2a = 6\), which means \(a = 3\).

For an ellipse, the relationship between \(a\), \(b\) (the length of the semi-minor axis), and \(c\) is given by \(c^2 = a^2 - b^2\).

We have \(c = 2\) and \(a = 3\), so \(2^2 = 3^2 - b^2\), which means \(4 = 9 - b^2\). Thus, \(b^2 = 9 - 4 = 5\), and \(b = \sqrt{5}\).

The length of the minor axis is \(2b = 2\sqrt{5}\).