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 $8x^2-y^2=$ $8a^2\text{ is }\frac{3}{2\sqrt{2}}$.

Equation of the normal at $(a\sec \theta ,b\tan \theta )$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

$\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2$

It meets the axes at $L\left(\frac{a^2+b^2}{a\cos \theta },0\right),M\left(0,\frac{a^2+b^2}{b\cot \theta }\right)$

$\text{ If }(h,k)\text{ is the mid-point of }LM$

$\text{ then }h=\frac{a^2+b^2}{2a\cos \theta },k=\frac{a^2+b^2}{2b\cot \theta }$

$\Rightarrow \sec \theta =\frac{2h}{9a},\tan \theta =\frac{2bk}{9a^2} \left[\because a^2+b^2=a^2{e}^2=9a^2\right]$

$\text{ Eliminating }\theta \text{, we get }$

$\frac{4h^2}{81a^2}-\frac{4b^2{k}^2}{81a^4}=1$

Locus of $(h,k)\text{ is }\frac{x^2}{\frac{81a^2}{4}}-\frac{y^2}{\frac{81a^4}{4b^2}}=1$

which is a hyperbola and its eccentricity

$\sqrt{1+\frac{81a^4}{4b^2}\times \frac{4}{81a^2}}=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+\frac{1}{e^2-1}}=\sqrt{1+\frac{1}{8}}$

$=\frac{3}{2\sqrt{2}}$

So statement-1 is true. In statement-2, the eccentricity  is

$\sqrt{1+\frac{8a^2}{a^2}}=3$ and the statement-2 is false