The normal at a variable point P on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, (a > b) meets axes of the ellipse in Q and R. Then which of the following is/are correct statement(s)

$\begin{array}{l}\text{ Equation of Normal to the Ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \text{ at }P(a\cos \theta ,b\sin \theta )\text{ is }\\ \frac{a^{2}x}{a\cos \theta }-\frac{b^{2}y}{b\sin \theta }=a^2-b^2\\ \frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2\\ \text{ it cuts }x\text{ -aris at }Q\\ Q=\left(\frac{\left(a^2-b^2\right)\cos \theta }{a},0\right)\\ \text{ and Normal cuts }y\text{ -axis at }R\\ R\left(0,-\frac{\left(a^2-b^2\right)\sin \theta }{b}\right)\\ \text{ Let }\left(x_1,y_1\right)\text{ be the mid point of }QR\\ \text{ Hence }{x}_1=\frac{\left(a^2-b^2\right)\cos \theta }{2a},y_1:-\frac{\left(a^2-b^2\right)\sin \theta }{2b}\\ \cos \theta =\frac{2a{x}_1}{a^2-b^2}\sin \theta =\frac{-2b{y}_1}{a^2-b^2}\\ \text{ Eliminate θ ,}\\ \frac{4a^2{x}_1^2}{{\left(a^2-b^2\right)}^2}+\frac{4b^2{y}_1^2}{{\left(a^2-b^2\right)}^2}=1\\ \frac{x_1^2}{\frac{{\left(a^2-b^2\right)}^2}{4a^2}}+\frac{y_1^2}{\frac{{\left(a^2-b^2\right)}^2}{4b^2}}=1\\ \text{ The locus of mid point is }\\ \frac{x^2}{{\displaystyle \frac{a^4{e}^4}{4a^2}}}+\frac{y^2}{{\displaystyle \frac{a^4{e}^4}{4b^2}}}=1\\ \text{ This represents Ellipse}\\ \text{ Eccentricity }e=\sqrt{1-\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a^2$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$b^2$}\right.}}}\\ =\sqrt{1-\frac{b^2}{a^2}}\\ \\ \text{ Same as that of given }Ellipse\end{array}$