Prove that the furthest point from the centre of ellipse is its vertex I got stuck a little bit with a simple proof (I think it must be simple but cannot figure it). I am quite certain these two should hold due to rigorously drawing the ellipse and the hyperbola ^^
Suppose we have only a vector definition of an ellipse. Given a unit normal vector $N$ to the directrix $L$, eccentricity $e$, and a set of points $X \in C$, where $C$ is an ellipse, we assume that the ellipse is symmetric around the origin, so $F = eaN$, where $a = \frac{ed}{1 - e^2}$. $F$ is one of the foci of the ellipse. Also $V = \pm aN$ are the vertices of the ellipse, while $\pm bN$ are the points connected by the minor axis (so the value $b$ can be used). The conic section is in the negative half-plane for simplicity. If we develop this further, we arrive at the ellipse vector equation:
$$\lVert X \rVert^2 + e^2a^2 = e^2(X \cdot N)^2  + a^2$$
Using this information, how do I actually show that $\lvert X \cdot N \rvert \leq a$. As I understood, this statement is close to saying that the vertex of the ellipse is the furthest point from the center among all $X \in C$. But somehow I cannot derive this from the above definitions. 
Can a similar fact be demonstrated for the hyperbola ($\lvert X \cdot N \rvert > |a|$)? 
 A: Another way
WLOG any point on $x^2/a^2+y^2/b^2=1,a\ge b$ can be $P(a\cos t,b\sin t)$ with $O(0,0)$ as center
$$OP^2=a^2\cos^2t+b^2\sin^2t=a^2-(a^2-b^2)\sin^2t\le a^2$$
Similarly for hyperbola,
$$OP^2=(a\sec t-0)^2+(b\tan t-0)^2=a^2+\tan^2t(a^2+b^2)\ge a^2$$
A: I could not find a suitable proof anywhere, so had to solve it myself. 
Given the equation of the conic section, presented in the question, we derive the necessary inequality $\lvert X \cdot N \rvert \leq a$ for the ellipse first, then derive the inequality $\lvert X \cdot N \rvert \geq |a|$ for the hyperbola. This is done with the help of the Cauchy–Schwarz inequality:
$$|X \cdot N| \leq \lVert X \rVert$$
Ellipse case. 
Given the Cauchy-Schwarz inequality, we infer
$$|X \cdot N| \leq \lVert X \rVert \quad \Rightarrow \quad \lVert X \rVert^2 + e^2a^2 \leq e^2\lVert X \rVert^2 + a^2$$
where we substituted $(X\cdot N)^2$ with $\lVert X \rVert^2$. Now we collect the vector length terms on one side, and $a$ term on the other side of the equation:
$$\lVert X \rVert^2 + e^2a^2 \leq e^2\lVert X \rVert^2 + a^2 \quad \Rightarrow \quad (1 - e^2)\lVert X \rVert^2 \leq (1 - e^2)a^2$$
But we know, that by the vector definition of conic sections, $\lVert X - F \rVert = ed(X, L)$, the eccentricity of the conic section, called ellipse is less than 1, $e < 1$. Thus, the term $1 - e^2$ is positive in the ellipse. We divide the both sides of the inequality by $1 - e^2$, and get the following: 
$$ \lVert X \rVert^2 \leq a^2 \quad \Rightarrow \quad \lVert X \rVert \leq a$$
Thus, we have shown that each point of the ellipse is closer to the origin than the vertex. As a consequence, by Cauchy-Schwarz, $\lvert X\cdot N\rvert \leq a$. 
Hyperbola case
Using the same Cauchy-Schwarz inequality, we substitute $\lVert X \rVert$ with $\lvert X\cdot N \rvert$ in the hyperbola equation:
$$|X \cdot N| \leq \lVert X \rVert \quad \Rightarrow \quad \lvert X\cdot N \rvert^2 + e^2a^2 \leq e^2|X \cdot N|^2 + a^2$$
Now grouping the dot-product terms on the right, and $a^2$ on the left:
$$(e^2 - 1)a^2 \leq (e^2 - 1)|X \cdot N|^2$$
But by the definition of the hyperbola as a conic section in vector form, its eccentricity is more than 1, $e > 1$. Hence, we can safely divide the both sides of the inequality with $e^2 - 1$, and arrive at $a^2 \leq |X \cdot N|^2$. From this we can of course see that $\lVert X \rVert \geq |X \cdot N| \geq |a|$, where $|a|$ is absolute, since $a$ is negative for the hyperbola (due to the definition of $a$ given in my question: denominator is negative in hyperbola).
Q.E.D.
PS. These proofs are necessary as a step in proofs for Apostol exercises 11 and 12, Section 13.25, Calculus 1 (section 13.25#11, section 13.25#12), so I am tagging it. 
