I have got a following problem :
$E = \{ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 \}$
$N$ - line (normal) perpendicular to E at $(x_0,y_0)$
Find max $dist(N,(0,0))$
So I am starting with attempt to find a tangent line to E at $(x_0,y_0)$. We shall call this line K.
$ K = \{ Df(x_0,y_0)(x-x_0,y-y_0) = 0\}$
$ ( \frac{2x}{a^2}(x_0,y_0), \frac{2y}{b^2}(x_0,y_0))(x-x_0,y-y_0)$ = 0
$ \frac{2x_0}{a^2}(x-x_0) + \frac{2y_0}{b^2}(y-y_0) = 0$
$ \frac{2xx_0}{a^2} - \frac{2(x_0)^2}{a^2} + \frac{2yy_0}{b^2} - \frac{2(y_0)^2}{b^2}=0$
So, after few transformations comes :
$y = - \frac{xx_0b^2}{a^2y_0} + \frac{(x_0)^2b^2}{a^2y_0} + y_0$
So, our line N is given by equation : $ N : y=cx+d$, $N \perp K$ at $(x_0,y_0)$
$- \frac{x_0b^2}{a^2y_0} \cdot c = -1$
$c = \frac{a^2y_0}{x_0b^2}$
$ N : y_0 = x_0 \cdot \frac{a^2y_0}{x_0b^2} +d $
$d = y_0 - \frac{a^2y_0}{b^2}$
So, our N is defined by $ y = \frac{a^2y_0}{x_0b^2}x + y_0 - \frac{a^2y_0}{b^2}$
distance from point to line is given by equation $$ dist(Ax+By+C,(x_1,y_1)) = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$$
So I get, hmmm
$$f(x_0,y_0) = \frac{|y_0 - \frac{a^2y_0}{b^2}|}{\sqrt {\frac{a^4(y_0)^2}{(x_0)^2b^4} +1}}$$
I get confused right here. I also tried to put new line coming out from (0,0), parallel to K and perpendicular to N $ L : y = - \frac{x_0b^2}{a^2y_0}x$.
When I tried to solve equation $$ - \frac{x_0b^2}{a^2y_0}x = \frac{a^2y_0}{x_0b^2}x + y_0 - \frac{a^2y_0}{b^2}$$ It wasn't any better. I am stuck.