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

Question

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 $Tt.PY=a^2-b^2$

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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 $\frac{x'-x_1}{A}=\frac{y'-y_1}{B}=\frac{-(Ax_1+By_1+C)}{A^2+B^2}$

Answer 1

It is the simplification of $PY$ that is not correct. However, the easy way is to note that

$$Tt\cdot PY=\overrightarrow{Tt}\cdot \overrightarrow{PC}= \left[\matrix{-a\sec\theta\cr b\csc\theta}\right]\cdot \left[\matrix{-a\cos\theta\cr -b\sin\theta}\right]=a^2-b^2.$$

Answer 2

Equation of $Tt$ is $bx\cos \theta + ay\sin \theta - ab = 0$. Hence $C{Y^2} = \frac{{ab}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$ and $C{P^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta $. Now, $$P{Y^2} = C{P^2} - C{Y^2} = \frac{{ab}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }} - \left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)$$ $$ = \frac{{{a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta - 1} \right) + {{\sin }^2}\theta {{\cos }^2}\theta \left( {{a^4} + {b^4}} \right)}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$ $$ = \frac{{{a^2}{b^2}\left( {{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}^2} - 2{{\sin }^2}\theta {{\cos }^2}\theta - 1} \right) + {{\sin }^2}\theta {{\cos }^2}\theta \left( {{a^4} + {b^4}} \right)}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$ $$ = \frac{{{a^2}{b^2}\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta - 1} \right) + {{\sin }^2}\theta {{\cos }^2}\theta \left( {{a^4} + {b^4}} \right)}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$ $$ = \frac{{{{\sin }^2}\theta {{\cos }^2}\theta {{\left( {{a^2} - {b^2}} \right)}^2}}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$ $$\Rightarrow PY = \frac{{\left( {{a^2} - {b^2}} \right)\sin \theta \cos \theta }}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}$$ Hence, $Tt \cdot PY=a^2-b^2.$