Question

Let \(x=-9\) be a directrix of an ellipse \(E\), whose centre is at the origin and eccentricity is \( \frac13 \). Let \(P(\alpha,0), \alpha>0\), be a focus of \(E\) and \(AB\) be a chord passing through \(P\). Then the locus of the mid point of \(AB\) is :

• A. \(9y^2=8x(1-x)\)
• B. \(3y^2=4x(1-x)\)
• C. \(9y^2=8x(x-1)\)
• D. \(3y^2=4x(x-1)\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We are given properties of an ellipse (directrix, eccentricity, center). We need to find the equation of the ellipse first. Then, we need to find the locus of the midpoint of a focal chord (a chord passing through a focus).
Step 2: Key Formula or Approach
For an ellipse centered at the origin \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \):
- Directrix: \( x = \pm \frac{a}{e} \).
- Focus: \( (\pm c, 0) \) where \( c = ae \).
- Relation: \( b^2 = a^2(1-e^2) = a^2 - c^2 \).
- Equation of a chord with midpoint \( (h,k) \) is given by \( T=S_1 \), where \( T = \frac{xh}{a^2} + \frac{yk}{b^2} - 1 \) and \( S_1 = \frac{h^2}{a^2} + \frac{k^2}{b^2} - 1 \).
Step 3: Detailed Explanation
Part 1: Find the equation of the ellipse.
Given directrix is \( x=-9 \). The formula for the directrix is \( x = -a/e \).
\[ \frac{a}{e} = 9 \] Given eccentricity \( e = 1/3 \).
\[ \frac{a}{1/3} = 9 \implies a = 3 \] The focus \( P(\alpha, 0) \) is at \( (\pm c, 0) \). Since \( \alpha>0 \), the focus is \( (c,0) \).
\[ c = ae = 3 \times \frac{1}{3} = 1 \] So the focus P is at \( (1,0) \).
Now, find \(b^2\). \[ b^2 = a^2 - c^2 = 3^2 - 1^2 = 9 - 1 = 8 \] The equation of the ellipse is: \[ \frac{x^2}{9} + \frac{y^2}{8} = 1 \] Part 2: Find the locus of the midpoint of the focal chord.
Let the midpoint of the chord AB be \( M(h,k) \).
The equation of the chord AB with midpoint \( (h,k) \) is \( T=S_1 \):
\[ \frac{xh}{9} + \frac{yk}{8} - 1 = \frac{h^2}{9} + \frac{k^2}{8} - 1 \] \[ \frac{xh}{9} + \frac{yk}{8} = \frac{h^2}{9} + \frac{k^2}{8} \] This chord passes through the focus \( P(1,0) \). So, we can substitute \(x=1, y=0\) into the chord's equation.
\[ \frac{(1)h}{9} + \frac{(0)k}{8} = \frac{h^2}{9} + \frac{k^2}{8} \] \[ \frac{h}{9} = \frac{h^2}{9} + \frac{k^2}{8} \] To get the locus, we rearrange this equation and replace \( (h,k) \) with \( (x,y) \).
\[ \frac{k^2}{8} = \frac{h}{9} - \frac{h^2}{9} \] \[ \frac{k^2}{8} = \frac{h(1-h)}{9} \] Multiply both sides by 72: \[ 9k^2 = 8h(1-h) \] Replacing \(h\) with \(x\) and \(k\) with \(y\), the locus is: \[ 9y^2 = 8x(1-x) \] Step 4: Final Answer
The locus of the midpoint of the chord AB is \( 9y^2=8x(1-x) \).