If normal at P to a hyperbola of eccentricity e intersects its transverse and conjugate axes at L and M, respectively, then prove that the locus of midpoint of LM is a hyperbola. Find the eccentricity of this hyperbola

Explanation

Consider hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1.`
Normal to any point `P(a sec theta, btan theta)` on the hyperbola is
`(ax)/(sec theta)+(by)/(tan theta)=a^(2)+b^(2)`
(2) It meets x-axis at `Q(((a^(2)+b^(2))sec theta)/(a),0)` and y-axis at `R(0,((a^(2)+b^(2))tantheta)/(b))`
Let midpoint of QR be T(h, k).
(3) Then `h=((a^(2)+b^(2))sec theta)/(2a)and k=((a^(2)+b^(2))tantheta)/(2b)`.
`therefore" "2h=ae^(2)secthetaand 2k=(a^(2)e^(2))/(b) tantheta`
`therefore" "(4h^(2))/(a^(2)e^(4))-(4k^(2))/(a^(4)e^(4))=1`
`rArr" "(x^(2))/((a^(2)e^(4))/4)-(y^(2))/((a^(4)e^(4))/(4b^(2)))=1,` which is required locus.
Clearly, it is hyperbola having eccentricity 'E' given by
`E^(2)=1+((a^(4)e^(2))/(4b^(2)))/((a^(2)e^(4))/(4))=1+(a^(2))/(b^(2))`
`=1+(1)/(e^(2)-1)`
`therefore" "E=sqrt((e^(2))/(e^(2)-1))=(2)/(sqrt3)" (as given that e = 2)"`