Product of distance and angles

[mathmari]

Hey!

I am looking at the following exercise:

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 $(\pm \epsilon p, 0)$ on the $x$-axis are called the foci of the ellipse, which we denote by $f_1$ and $f_2$. Verify that $\gamma (t) = (p \cos t, q \sin t)$ is a parametrization of the ellipse. Prove that:

1. The product of the distances from $f_1$ and $f_2$ to the tangent line at any point $p$ of the ellipse does not depend on $p$.
   
2. 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 $p$.

Since $\gamma (t)=(p \cos t , q \sin t)$ we have that $\gamma '(t)=(-p\sin t, q\cos t)$.

The tangent line is given by $l(t)=\gamma (t_0)+t\gamma' (t_0)$.

How can we calculate the distances from $f_1$ and $f_2$ to the tangent line?

Could you give me some hints what we could do at the second question?

 

[I like Serena]

Since $\gamma (t)=(p \cos t , q \sin t)$ we have that $\gamma '(t)=(-p\sin t, q\cos t)$.

The tangent line is given by $l(t)=\gamma (t_0)+t\gamma' (t_0)$.

How can we calculate the distances from $f_1$ and $f_2$ to the tangent line?

Could you give me some hints what we could do at the second question?

Hi mathmari! (Wave)

The distance of a point $\mathbf F$ to a line $l$ through $\mathbf P$ in the direction $\mathbf d$, is given by:
$$d(\mathbf F, l) = ||\boldsymbol\pi_{\mathbf n}(\mathbf F-\mathbf P)|| = (\mathbf F-\mathbf P) \cdot \mathbf n$$
where $\boldsymbol\pi_{\mathbf n}$ is the projection on a normal $\mathbf n$ of the line (perpendicular to $\mathbf d$).
And that projection is given by the dot product if $\mathbf n$ is a vector of unit length.

In your case $\mathbf n$ is given by $(q\cos t, p\sin t)$.

For the second question I think you're supposed to use another application of the dot product.

 

[math771]

Hey there!

For the first question, we can use the distance formula to calculate the distances from $f_1$ and $f_2$ to the tangent line. Let's say the point $p$ on the ellipse has coordinates $(x_0, y_0)$. Then, the distance from $f_1$ to the tangent line at $p$ would be:

$$d_1 = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$

where $ax + by + c = 0$ is the equation of the tangent line at $p$. Similarly, the distance from $f_2$ to the tangent line at $p$ would be:

$$d_2 = \frac{|ax_0 + by_0 + c'|}{\sqrt{a^2 + b^2}}$$

where $ax + by + c' = 0$ is the equation of the other tangent line at $p$. Now, we can take the product of these distances and show that it does not depend on $p$.

For the second question, we can use the fact that the tangent line is perpendicular to the radius vector at the point $p$. This means that the slopes of the lines joining $f_1$ and $p$, and $f_2$ and $p$ to the tangent line at $p$ will be equal. From this, we can show that the angles made by these lines with the tangent line will also be equal.

Hope this helps! Let me know if you have any other questions.