The normal at `P` to a hyperbola of eccentricity `e`, intersects its transverse and conjugate axes at `L` and `M` respectively. Show that the locus of the middle point of `LM` is a hyperbola of eccentricity `e/sqrt(e^2-1)`

Options

A. `(e+1)/(e-1)`
B. `(e)/(sqrt(e^(2)-1)`
C. `e`
D. none of these

Explanation

To solve the problem step by step, we will derive the locus of the midpoint of the segment \(LM\) formed by the intersection of the normal at point \(P\) to a hyperbola with its transverse and conjugate axes.

### Step 1: Define the Hyperbola
The standard equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.
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