Being inside or outside of an ellipse Let $A$ be a point $A$ not belonging to an ellipse $E$. We say that $A$ lies inside
$E$ if every line passing trough $A$ intersects $E$. We say that $A$ lies otside $E$
if some line passing trough $A$ does not intersect $E$. Let $E$ be the ellipse
with semi-axes $a$ and $b$. Show that
(a) a point $A=(x,y)$ is inside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}< 1$
(b) a point $A=(x,y)$ is outside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}> 1$.
I wrote down the equation of an arbitrary line $L$ passing through $A$. If $A$ is inside $E$ then clearly there is another point, say $B$, that belongs to $A\cap L$. I ended up with a bunch of big equations which didn't give any result. 
 A: Set the line $L$ be passing through the origin and a point $A(x_0,y_0),x_0\neq0$ inside the ellipse. So we have $$L: y=\frac{y_0}{x_0}x$$ This line should intersect the ellipse in another point(s), so we should consider two equations below simultaneously: $$y=\frac{y_0}{x_0}x,~~~ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Then we get $$\left(\frac{y_0^2}{b^2x_0^2}+\frac{1}{a^2}\right)x^2=1~~~~\text{or}~~~~~\frac{y_0^2}{b^2}+\frac{x_0^2}{a^2}=\frac{x_0^2}{x^2}$$ But $\frac{x_0^2}{x^2}<1$.
A: Case $a:$
Let the point be $P(h,k)$ and any arbitrary line passing through $P$ be $y=mx+c$
$\implies k=m\cdot h+c\iff c=k-m\cdot h$
Let us find the intersection of the line with the given ellipse
$${So,}\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\implies \dfrac{x^2}{a^2}+\dfrac{(mx+c)^2}{b^2}$$
$$\text{On simplification,}x^2(b^2+m^2)+2a^2cm x+a^2(c^2-b^2)=0$$
For intersection, there will be $2$ distinct values of $m$
i.e., the discriminant $(2a^2cm)^2-4(b^2+m^2)a^2(c^2-b^2)=4a^2(a^2m^2+b^2-c^2)>0$
i.e., $a^2m^2+b^2-c^2>0$
Putting the value of $c, a^2m^2+b^2-(k-m\cdot h)^2>0$
Putting $m=0, b^2-k^2>0$
Put $m=\frac1n, \frac{a^2}{n^2}+b^2-(k- \frac hn)^2>0 \iff a^2+b^2n^2-(k\cdot n-h)^2>0$
Putting $n=0, a^2-h^2>0$
$$\implies (b^2-k^2)(a^2-h^2)>0 \implies a^2b^2> a^2k^2+ b^2h^2\implies 1> \frac{k^2}{b^2}+\frac{h^2}{a^2}$$
Case $b:$ there will be no intersection i.e.,  the discriminant $<0$ 
