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 :
- \(9y^2=8x(1-x)\)
- \(3y^2=4x(1-x)\)
- \(9y^2=8x(x-1)\)
- \(3y^2=4x(x-1)\)
The Correct Option is A
Solution and Explanation
Concept: For an ellipse centered at origin: \[ e=\frac{c}{a}, \qquad \text{directrix } x=\pm\frac{a}{e} \] The midpoint of a focal chord satisfies a standard locus relation derived from parametric representation of the ellipse.
Step 1: {Find semi-major axis.} Given: \[ \frac{a}{e}=9 \] \[ \frac{a}{1/3}=9 \] \[ 3a=9 \] \[ a=3 \]
Step 2: {Find \(b\).} \[ c=ae=1 \] \[ b^2=a^2-c^2 \] \[ b^2=9-1 \] \[ b^2=8 \] Ellipse equation: \[ \frac{x^2}{9}+\frac{y^2}{8}=1 \]
Step 3: {Use midpoint of focal chord property.} For an ellipse, the locus of midpoint of focal chords reduces to: \[ 9y^2=8x(1-x) \]
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