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The abscissae of A and B on the circle $x^{2} + y^{2} = 25$ are respectively 5 and –5. Let MN be a variable chord perpendicular to AB, and if the locus of point of intersection of the lines $\bar{BM}$ and $\vec{AN}$ is a conic then its eccentricity is

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$\frac{1}{5}$

$\sqrt{5}$

$\sqrt{2}$

answer is B.

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Detailed Solution

Let $M = (5 \cos θ, 5 \sin θ)$

$N = (5 \cos θ, - 5 \sin θ)$

Slope $BM = \frac{\sin θ}{1 + \cos θ}$

Slope $NA = \frac{\sin θ}{1 - \cos θ}$

$\begin{array}{l} \vec{BM} : y = \frac{\sin θ}{1 + \cos θ} (x + 5) \\ \vec{AN} : y = \frac{\sin θ}{1 - \cos θ} (x - 5) \begin{array}{l} \end{array} \end{array}$

Solving : $Point of intersection of BM, AN = P : (5 \sec θ, 5 \tan θ)$

Locus is $x^{2} - y^{2} = 25, e = \sqrt{2}$

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