why such a point exists? Given oval $$O: \frac{x^2}6+\frac{y^2}3=1$$ and point $A (2,1)$ on it, draw two lines $l_1$, $l_2$ passing through  $A$ s.t. the slopes satisfy $k_1k_2=2$, let $l_1$ and $l_2$ intersect with $O$ at $B$ and $C$, then line $BC$ always pass point $D(\frac{10}3, -\frac{5}3)$. 
My question is, why such a point $D$ exists? what's the relationship between $A$ and $D$?
 A: We can scale the problem by taking $y\to y\sqrt{2}$. Then
$$A \to (2,\sqrt{2}),$$
$$O \to x^2+y^2=6,$$
and the condition that $k_1k_2=2$ is replaced by the condition that $k_1k_2=4$. 
We claim the following:

Let $A$ be a point on circle $\Omega$ centered at point $O$, let $P$ be the reflection of $A$ about the $x$-axis, and let $t$ be a real number. Let $Q$ be the second intersection of the line through $A$ with slope $t$ with $\Omega$, and let $B$ and $C$ be points on $O$ such that the slopes of lines $AB$ and $AC$ multiply to $t^2$. Then lines $OP$, $BC$, and the tangent to $\Omega$ at $Q$ concur.

This shouldn't be too hard to prove, though I don't see how to do it immediately. But, to answer your second question ("What's the relationship between $A$ and $D$?), $D$ is the intersection between the line $OP$ (the reflection of line $OA$ across the $x$-axis) and the tangent line that the line $BC$ tends to when $B\to C$. 
Once can see that this should be what $D$ is, if $D$ does indeed exist, because the line $OP$ is what $BC$ tends to when one of ($AB$, $AC$) has slope $\epsilon$ and the other has slope $t^2/\epsilon$ (one line tends to vertical and the other tends towards horizontal), while the tangent at $Q$ to $\Omega$ is what $BC$ tends to when $B\to C$. 
A: This fact is true for any point $A$ of the ellipse and any other value for the constant $2$.
For every point $B$, the maps
$$ point(B) \mapsto line(AB) \mapsto line(AC) \mapsto point(C) $$
are projective, so their composition $B\mapsto C$ is a projective transform from the ellipse to itself. Moreover, $C$ is mapped back to $B$, so this transform is an involution.
Involutions are always projections from a certain point which is $D$ in our case.
A: Let $O$ be an ellipse, and fix some $D \notin O$.
Then there is a natural involution $i_D : O \to O$ from the ellipse to itself where a point $P \in O$ is sent to the other intersection point of $(DP)$ with $O$. Furthermore, this involution is algebraic : the coordinates of $i_D(P)$ are some quotient of polynomials in terms of the coordinates of $P$.
Then, fix some $A \in O$ and let $s_A : O \to \Bbb P^1(\Bbb R) = \Bbb R \cup \{ \infty \}$ be the function  mapping a point $P$ to the slope of $(AP)$. This is also an algebraic map.
Finally let $h(s) = 2/s$ be the automorphism of $\Bbb P^1(\Bbb R)$ that you're looking at.
If you pick two random points $B_1,B_2$ on $O$, your construction sends them to $C_1,C_2$ where $C_i = s_A^{-1} \circ h \circ s_A (B_i)$. Then, if you take $D$ to be the intersection of $(B_1C_1)$ and $(B_2C_2)$ then $i_D \circ s_A^{-1} \circ h \circ s_A$ is an automorphism of $O$ that has at least four fixed points.
Or rather, $s_A \circ i_D \circ s_A^{-1} \circ h$ is an automorphism of $\Bbb P^1(\Bbb R)$ that has at least four fixed points.
But what can this algebraic map be ? Since it's algebraic it's a map $F : s \mapsto f(s)/g(s)$ where $f,g$ are polynomials. Moreover since it is injective, $f$ and $g$ must have degree at most $1$, so $F(s)$ has the form $(as+b)/(cs+d)$.
Then $F(s)-s = (-cs^2+(a-d)s+b) / (cs+d)$ and this has at least four zeroes. But its numerator only has degree $2$, so $F(s)-s$ must be identically zero, which means that $s_A \circ i_D \circ s_A^{-1} \circ h$ (and $i_D \circ s_A^{-1} \circ h \circ s_A$) are the identity.

Or if you want to say all this really concisely, an ellipse is a genus $0$ curve (like $\Bbb P^1(\Bbb R)$) and so its automorphisms are determined by the images of $3$ points. Since $i_D$ and $s_A^{-1} \circ h \circ s_A$ agree on $4$ points, they must match completely.
