If the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively If  the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively and $CY$ is perpendicular on the tangent from the center,then prove that $Tt.PY=a^2-b^2$

Let the point $P$ be $(a\cos\theta,b\sin\theta)$,then the  equation of the tangent  to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  is $\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$.
It meets the major axis at $T(a\sec\theta,0)$ and the minor axis at $t(0,b\csc\theta)$
$Tt=\sqrt{a^2\sec^2\theta+b^2\csc^2\theta}$ 
Now I found $Y$,the foot of perpendicular from center $C$ of the ellipse to the tangent as $(\frac{\frac{\cos\theta}{a}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}},\frac{\frac{\sin\theta}{b}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}})$
$PY=\sqrt{(a\cos\theta-\frac{\frac{\cos\theta}{a}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}})^2+(b\sin\theta-\frac{\frac{\sin\theta}{b}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}})^2}$
I simplified this expression to get $PY=\frac{(a^2-b^2)\sin\theta\cos\theta}{(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2})a^2b^2}\sqrt{a^4\sin^2\theta+b^4\cos^2\theta}$ and $Tt=\sqrt{a^2\sec^2\theta+b^2\csc^2\theta}=\frac{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}{\sin\theta \cos\theta}$
But $PY.Tt$ is not $a^2-b^2$I do not understand where have i gone wrong.Maybe i found $Y$ wrong,i used the formula for the foot of the perpendicular $(x',y')$ from $(x_1,y_1)$ on the line $Ax+By+C=0$ as $\frac{x'-x_1}{A}=\frac{y'-y_1}{B}=\frac{-(Ax_1+By_1+C)}{A^2+B^2}$ 

It is the simplification of $PY$ that is not correct. However, the easy way is to note that
$$Tt\cdot PY=\overrightarrow{Tt}\cdot \overrightarrow{PC}=
\left[\matrix{-a\sec\theta\cr b\csc\theta}\right]\cdot
\left[\matrix{-a\cos\theta\cr -b\sin\theta}\right]=a^2-b^2.$$
