Question

One of the foci of an ellipse is \( (2,-3) \) and its corresponding directrix is \( 2x+y=5 \). If the eccentricity of the ellipse is \( \frac{\sqrt{5}}{3} \) then the coordinates of the other focus are

• A. \( (18, 5) \)
• B. \( (4, -2) \)
• C. \( (-2, -5) \)
• D. \( (-4, -6) \)

Hint:

Using the relation \( 9\vec{S} = 4\vec{C} + 5\vec{Z} \) (derived from section formula) is more efficient than distance equations involving variables.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We find the center of the ellipse first. The center lies on the major axis (line passing through the focus and perpendicular to the directrix). We can determine the center's location using the relationship between the focus, center, and directrix distances. Once the center is found, the other focus can be found using the midpoint formula.
Step 2: Key Formula or Approach:

1. Distance from center to focus = \( ae \). 2. Distance from center to directrix = \( a/e \). 3. Section formula or vector approach to find the center.
Step 3: Detailed Explanation:

Let \( S(2, -3) \) be the focus and the directrix be \( 2x+y-5=0 \). Let \( Z \) be the foot of the perpendicular from \( S \) to the directrix. Distance \( SZ \) (perpendicular distance from S to line): \[ SZ = \frac{|2(2) + (-3) - 5|}{\sqrt{2^2 + 1^2}} = \frac{4}{\sqrt{5}} \] On the major axis, the points are ordered as Directrix -- Center -- Focus -- Vertices? No, typically Center \( C \), Focus \( S \) at \( ae \), Directrix \( Z \) at \( a/e \). Distance \( CS = ae \). Distance \( CZ = a/e \). Distance \( SZ = CZ - CS = \frac{a}{e} - ae = a\left(\frac{1}{e} - e\right) \). Substitute \( e = \frac{\sqrt{5}}{3} \): \[ \frac{4}{\sqrt{5}} = a \left( \frac{3}{\sqrt{5}} - \frac{\sqrt{5}}{3} \right) = a \left( \frac{9-5}{3\sqrt{5}} \right) = \frac{4a}{3\sqrt{5}} \] \[ \frac{4}{\sqrt{5}} = \frac{4a}{3\sqrt{5}} \implies a = 3 \] Now find distances from Center \( C \): \( CS = ae = 3(\frac{\sqrt{5}}{3}) = \sqrt{5} \). \( CZ = a/e = 3(\frac{3}{\sqrt{5}}) = \frac{9}{\sqrt{5}} = 1.8\sqrt{5} \). Since \( S \) lies between \( C \) and \( Z \), \( S \) divides \( CZ \) in the ratio \( CS : SZ = \sqrt{5} : \frac{4}{\sqrt{5}} = 5 : 4 \). Wait, \( CZ \textgreater CS \), so \( S \) is between \( C \) and \( Z \). \( \vec{S} = \frac{4\vec{C} + 5\vec{Z}}{9} \implies 9\vec{S} = 4\vec{C} + 5\vec{Z} \implies 4\vec{C} = 9\vec{S} - 5\vec{Z} \). Find coordinates of \( Z \): Axis line equation (perp to \( 2x+y=5 \) through \( (2,-3) \)): \( x - 2y = 2 - 2(-3) = 8 \). Solve \( 2x+y=5 \) and \( x-2y=8 \): Multiply first by 2: \( 4x+2y=10 \). Add to second: \( 5x=18 \implies x=3.6 \). \( y = 5 - 2(3.6) = -2.2 \). \( Z(3.6, -2.2) \). Calculate \( C \): \( 4x_C = 9(2) - 5(3.6) = 18 - 18 = 0 \implies x_C = 0 \). \( 4y_C = 9(-3) - 5(-2.2) = -27 + 11 = -16 \implies y_C = -4 \). Center \( C(0, -4) \). Find other focus \( S' \): \( C \) is midpoint of \( S \) and \( S' \). \( \frac{2 + x'}{2} = 0 \implies x' = -2 \). \( \frac{-3 + y'}{2} = -4 \implies y' = -5 \). \( S'(-2, -5) \).
Step 4: Final Answer:

The coordinates are \( (-2, -5) \).