Question

If the focus of an ellipse is \((-1,-1)\), equation of its directrix corresponding to this focus is \(x + y + 1 = 0\) and its eccentricity is \(\frac{1}{\sqrt{2}}\), then the length of its major axis is:

• A. 2
• B. 1
• C. 4
• D. 3

Hint:

- When working with ellipses, remember the relationship between the eccentricity, the distance from the center to the focus (\(c\)), and the semi-major axis (\(a\)). Use the focus-directrix property to find the necessary lengths.

Answer 1

The Correct Option is A

The eccentricity of an ellipse is given by the formula: \[ \epsilon = \frac{c}{a} \] where \(c\) is the focal length (the distance from the center to the focus) and \(a\) is the semi-major axis length (the distance from the center to the vertices along the major axis). Given that the eccentricity is \(\frac{1}{\sqrt{2}}\), we have: \[ \epsilon = \frac{1}{\sqrt{2}} \] The distance from the focus to the directrix is given by the equation for the focus-directrix property of the ellipse: \[ \text{Distance from focus to directrix} = \frac{1}{\epsilon} \] Substituting the value of \(\epsilon\): \[ \text{Distance from focus to directrix} = \sqrt{2} \] Now, we calculate the perpendicular distance from the focus \((-1, -1)\) to the directrix \(x + y + 1 = 0\). Using the formula for the perpendicular distance from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\): \[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] For the directrix \(x + y + 1 = 0\), \(A = 1\), \(B = 1\), and \(C = 1\), and the point is \((-1, -1)\). Substituting into the distance formula: \[ \text{Distance} = \frac{|1(-1) + 1(-1) + 1|}{\sqrt{1^2 + 1^2}} = \frac{|-1 - 1 + 1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \] Thus, the distance from the focus to the directrix is \(\frac{1}{\sqrt{2}}\), which confirms that the semi-major axis length \(a = \sqrt{2}\). Finally, the length of the major axis is \(2a\), so: \[ 2a = 2 \times \sqrt{2} = 2 \] Thus, the length of the major axis is \(\boxed{2}\).