Question involving gradient of a function.

Question

We are given any arbitrary ellipse with focii $F1$ and $F2$ , $T$ is the unit tangent to the ellipse through a point $P$. Let $f$ be the sum of the distances of the of $F1$ and $F2$ from $P$ , we need to show that $T.\nabla f=0$ , also using this we need to show the angles $a$ and $b$ are equal.

What I tried :

We know that the sum of distances from $P$ to two fixed points (focii) is a constant , so we can let $f=k$ , where $k$ is any constant.

Now , $\nabla f = f_x i + f_y j$ , => $\nabla f = 0$ ,

Now we write the equation of an ellipse as follows :

$R(\theta) = acos\theta i + asin\theta j$ ( I guess we can write the equation like this).

So tangent is given by : $R^{'}(\theta) = -asin\theta i + bcos\theta j$ ,

So a unit vector $T$ would be : $ \dfrac{-asin\theta i + acos\theta j}{\sqrt{a^{2}sin^{2}\theta + b^{2}cos^{2}\theta}}$

Clearly , $T.\nabla f=0$ , Is this OK ?

And I have no idea how to use this to prove $a=b$ , could anyone help ?

Answer 1

We may assume that the two foci lie at $(-c,0)$ and $(c,0)$ and the length of the major semi-axis is $a$ as usual. Then a point $P=(x,y)$ on the ellipse fulfills:

$$ \sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2} = 2a \tag{1}$$ and by implicit differentiation we get: $$ \frac{(x+c)dx+y\,dy}{\sqrt{(x+c)^2+y^2}}+\frac{(x-c)dx+y\,dy}{\sqrt{(x-c)^2+y^2}}=0.\tag{2}$$ On the other hand, what is the intersection $Q$ of the $F_1 F_2$-line with the angle bisector of $\widehat{F_1 P F_2}$?

By the angle bisector theorem we have: $$ \frac{F_1 Q}{Q F_2} = \frac{\sqrt{(x+c)^2+y^2}}{\sqrt{(x-c)^2+y^2}} \tag{3}$$ hence $Q$ lies at $\left(c\cdot \frac{\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}}{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}},0\right)$ and it is not difficult to compute the equation of the $PQ$-line. On the other hand, $(2)$ gives the equation of the tangent $t_P$ to the ellipse through $P$. By computing a dot product, it is straightforward to check that $t_P\perp PQ$ as wanted.

Answer 2

First of all, I'm afraid that your argument is incorrect, because $\nabla f$ is not identically zero. The problem is your starting point: for $\nabla f$ to be identically zero, $f$ has to be identically a constant (on $\mathbb R^2$, or in some neighborhood of the ellipse, at least); $f$ is constant on the ellipse - which is of course what you are thinking - but that is not enough to make the gradient identically zero, even on the ellipse. But maybe you are just messing up the notation?

In any event:

Suppose $R \colon \mathbb R \rightarrow \mathbb R^2$ is a parametrization of the ellipse such that $R ( 0 ) = P$ . Then $$ f ( R ( t ) ) = k,$$ by the definition of $f$ and of the ellipse ($k$ is the length of the major axis). Take the derivative of both sides of the equation. We get, from the chain rule on the left hand side of the equation, and from $k$ being a constant on the right, that
$$ (\nabla f )_{R ( t) } \cdot R'(t) = 0, $$ where $ (\nabla f )_{R ( t) } $ means $\nabla f$ evaluated at the point $R(t)\in \mathbb R^2$. If we choose $R$ to be parametrization by arc length, then $R'(0)$ is a unit tangent vector $T$ at $P$: $$ (\nabla f )_P \cdot T = 0.$$

The second part of the question:

We have that $$ \cos b = { {(F_2 -P ) \cdot T} \over |F_2 -P |}$$ and $$ \cos a = { - {(F_1 -P ) \cdot T} \over |F_1 -P |}.$$ This is because $u \cdot v =|u| \ |v| \cos \theta$ for any $u$ and $v$ vectors, and $\theta$ is the angle between them. Here, $T$ has length $1$, so doesn't appear in the denominator, and there is a minus sign in front of the second equation because we are using $-T$.

Let $l_i (t ) = | R(t)- F_i|$. Then $$ l_1(t) + l_2(t) = k,$$ so that, differentiating on both sides, $$ l_1'(t) + l_2'(t) = 0. $$ Therefore $$l_1'(t)=- l_2'(t).$$ Now, if
$$ l_i' (t) = {(R(t) - F_i ) \cdot T \over |R(t) - F_i |}, $$ we are done - the two angles are the same.

But that is the case: Say $d(t)$ is the distance between $R(t)$ and some fixed point $A$. Then, $d(t) = | R(t) - A|$, and
$$[ d(t) ]^2 = (R(t) -A ) \cdot (R(t) -A) . $$ Differentiating both sides, and using the chain rule on the left, and the fact that $(u\cdot v)'= u'\cdot v + u\cdot v'$ and the commutativity of the dot product on the right, so that $(u\cdot u)' = 2u\cdot u'$, we get $$2 d(t) d'(t) = 2 (R(t) -A ) \cdot R'(t), $$ so that $$ d'(t) = {(R(t) -A ) \cdot R'(t)\over d(t)},$$ and we are done, since $ d(t) = | R(t) - A|$.