Statement 1: A normal to the hyperbola with eccentricity 3, meets the transverse axis and conjugate axis at $P$ and $Q$ respectively. The locus of the mid-point of 

$PQ$ is a hyperbola with eccentricity $\frac{3}{2\sqrt{2}}$

Statement 2: Eccentricity of the hyperbol $8x^2-y^2=$ $8a^2\text{ is }\frac{3}{2\sqrt{2}}$.

detailed solution

Correct option is C

Equation of the normal at (asec⁡θ,btan⁡θ) to the hyperbola x2a2−y2b2=1 isaxsec⁡θ+bytan⁡θ=a2+b2It meets the axes at La2+b2acos⁡θ,0,M0,a2+b2bcot⁡θ If (h,k) is the mid-point of LM then h=a2+b22acosθ,k=a2+b22bcotθ⇒secθ=2h9a,tanθ=2bk9a2  ∵a2+b2=a2e2=9a2 Eliminating θ, we get 4h281a2-4b2k281a4=1Locus of (h,k) is x281a24-y281a44b2=1which is a hyperbola and its eccentricity1+81a44b2×481a2=1+a2b2=1+1e2-1=1+18=322So statement-1 is true. In statement-2, the eccentricity  is1+8a2a2=3 and the statement-2 is false