We consider the ellipse $$\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$$ where $p>q>0$. The eccentricity of the ellipse is $\epsilon =\sqrt{1-\frac{q^2}{p^2}}$, and the points $(\pm \epsilon p, 0)$ of the axis $x$ are called focal points and are symbolized with $F_1$ and $F_2$.
Verify that $\gamma (t)=(p \cos t, q \sin t)$ is a parametrization of the ellipse.
Prove that the sum of the distance of a point $K$ of the ellipse from $F_1$ and $F_2$ doesn't depend on $K$.
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I have done the following:
For $x=p \cos t$ and $y=q \sin t $ we have $$\frac{x^2}{p^2}+\frac{y^2}{q^2}=\frac{p^2\cos^2 t}{p^2}+\frac{q^2 \sin^2 t}{q^2}=\cos^2 t+\sin^2 t=1$$
So $\gamma (t)$ is a parametrization of the ellipse.
Let $K=(x, y)$. Let $D_1$ be the distance of $K$ from $F_1=(\epsilon p, 0)$ and $D_2$ the distance of $K$ from $F_2=(-\epsilon p, 0)$.
Then $D_1^2=(x-\epsilon p)^2+y^2$ and $D_2^2=(x+\epsilon p)^2+y^2$.
For $x=p\cos t , y=q\sin t$ we have $$D_1^2=(p\cos t-\epsilon p)^2+q^2\sin^2 t = \dots =p^2(1-\cos t \epsilon )^2 \Rightarrow D_1=p(1-\cos t\epsilon) \\ D_2^2=(p\cos t+\epsilon p)^2+q^2\sin^2 t = \dots =p^2(1+\cos t \epsilon )^2 \Rightarrow D_2=p(1+\cos t\epsilon)$$ So $$D_1+D_2=2p$$ Is this correct? Could I improve something at the fomulation?
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EDIT1:
After that I am asked to prove that the product of the distances of the tangent line at a point $L$ of the ellipse from $F_1$ and $F_2$ doesn't depend on $L$.
I have found the tangent vector $\gamma '(t)=(-p \sin t, q \cos t)$.
What do I have to do next? I got stuck right now...
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EDIT2:
If $p$ is any point on the ellipse, the line joining $f_1$ and $p$ and that joining $f_2$ and $p$ make equal angles with the tangent line to the ellipse at $p$.
Could you give me some hints what we are supposed to do in this case?
Vectors are not very useful when you want to calculate how far a point is from a line, because vectors are not localised. So I use cartesian coordinates instead.
The distance of a point $(x_0,y_0)$ from a line $ax+by+c=0$ is given by the formula $$\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$
For your question, you can take the equation of the tangent as $qx\cos\theta+py\sin\theta=pq$ , and the problem is then pretty straightforward.
EDIT: This is to answer your edit.
Say the foot of perpendiculars dropped from the foci $F_1$ and $F_2$ on the tangent at $P$ are $S_1$ and $S_2$. Using the distance formula above you can easily find the ratio $\dfrac{F_1S_1}{F_2S_2}$. Using the definition of ellipse it is very easy to find the ratio $\dfrac{PF_1}{PF_2}$ . Show that these ratios are equal and hence you'll prove that $\triangle PS_1F_1 \sim \triangle PS_2F_2$