Question

An ellipse has directrix \(x = 9\) & eccentricity \(= \frac{1}{3}\). If one of its focus is \((\alpha,0)\), \(\alpha<0\), then locus of the mid-point of the chord passing through \(P(\alpha,0)\) is

• A. \(\frac{x^2}{9} + \frac{y^2}{8} = \frac{x}{9}\)
• B. \(\frac{x^2}{9} + \frac{y^2}{2} = \frac{x}{9}\)
• C. \(\frac{x^2}{8} + \frac{y^2}{2} = \frac{x}{8}\)
• D. \(\frac{x^2}{9} + \frac{y^2}{8} = \frac{x}{4}\)

Hint:

For any conic \(S=0\), the locus of the mid-point of chords passing through a fixed point \((x_1, y_1)\) is always obtained by substituting the point into the equation \(T = S_1\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We first find the equation of the ellipse. Then we use the mid-point chord formula \(T = S_1\) and pass it through the given focus.

Step 2: Key Formula or Approach:
1. Ellipse parameters: \(a/e = 9 \implies a = 9(1/3) = 3\). 2. \(b^2 = a^2(1 - e^2) = 9(1 - 1/9) = 8\). 3. Focus: \((ae, 0) = (3 \cdot 1/3, 0) = (1, 0)\). So \(\alpha = 1\).

Step 3: Detailed Explanation:
1. Ellipse Equation: \(\frac{x^2}{9} + \frac{y^2}{8} = 1\). 2. Locus of mid-point \((h, k)\): The equation of the chord is \(T = S_1\). \[ \frac{xh}{9} + \frac{yk}{8} = \frac{h^2}{9} + \frac{k^2}{8} \] 3. Since the chord passes through focus \((1, 0)\): \[ \frac{(1)h}{9} + \frac{(0)k}{8} = \frac{h^2}{9} + \frac{k^2}{8} \] \[ \frac{h}{9} = \frac{h^2}{9} + \frac{k^2}{8} \] 4. Replace \((h, k)\) with \((x, y)\): \[ \frac{x^2}{9} + \frac{y^2}{8} = \frac{x}{9} \]

Step 4: Final Answer:
The locus of the mid-point is \(\frac{x^2}{9} + \frac{y^2}{8} = \frac{x}{9}\).