Longest distance of a point from an ellipse So here is a question I was trying to solve: The longest distance of the point $(a,0) $ from the curve $2x^2+y^2-2x=0$ is given by ?
My Attempt: The curve in the question represents an ellipse. So I wrote the parametric form of the ellipse to get the co ordinates of any random point on the ellipse. Then I wrote the expression for distance of the point $(a,0) $ from that random point and applied maxima minima to get the longest distance. 
This method does work but is a bit lengthy. The second method I found was using Langrange Multipliers. But it is also lengthy. Is there any direct result for such a question? Like I know that shortest distance of a point from a curve is along the common normal. Does anything like that exist for longest distance?
Note: A similar question exists here
But my question is a bit different. The point exists on X axis and is along the minor axis of ellipse. Maybe that could help simplify the things. 
 A: Note that
$$
\eqalign{
  & 0 = 2x^{\,2}  - 2x + y^{\,2}  = 2\left( {x^{\,2}  - x} \right) + y^{\,2}  =   \cr 
  &  = 2\left( {x^{\,2}  - x + {1 \over 4} - {1 \over 4}} \right) + y^{\,2}  = 2\left( {\left( {x - {1 \over 2}} \right)^{\,2}  - {1 \over 4}} \right) + y^{\,2} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad {{\left( {x - {1 \over 2}} \right)^{\,2} } \over {\left( {1/2} \right)^{\,2} }} + {{y^{\,2} } \over {\left( {\sqrt 2 /2} \right)^{\,2} }} = 1 \cr} 
$$
so there is no need of invoking sophisticated methods.
The above canonical equation of the ellipse is telling you that it is centered at (1/2,0),
 has the $x$ semi-axis of $1/2$, so the distance from it of a point $A=(a,0)$ 
will be when the circle centered in $A$ will have two coincident intersections with a proper quarter of the ellipse.
Now, as the sketch shows, for $A$ internal to the ellipse we may have up to three tangential circle centered on $A$.

We have to choose the circle which provides $0<y_T$. Therefore
$$
\eqalign{
  & \left\{ \matrix{
  \left( {x - a} \right)^{\,2}  + y^{\,2}  - R^{\,2}  = 0 \hfill \cr 
  2\left( {x^{\,2}  - x} \right) + y^{\,2}  = 0 \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  0 < y = \sqrt {R^{\,2}  - \left( {x - a} \right)^{\,2} }  \hfill \cr 
  2\left( {x^{\,2}  - x} \right) + R^{\,2}  - \left( {x - a} \right)^{\,2}  = 0 \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  0 < y = \sqrt {R^{\,2}  - \left( {x - a} \right)^{\,2} }  \hfill \cr 
  x^{\,2}  - 2\left( {1 - a} \right)x + R^{\,2}  - a^{\,2}  = 0 \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  0 < y = \sqrt {R^{\,2}  - \left( {x - a} \right)^{\,2} }  \hfill \cr 
  x_{\,1,2}  = \left( {1 - a} \right) \pm \sqrt {\left( {1 - a} \right)^{\,2}  - R^{\,2}  + a^{\,2} }  \hfill \cr 
  0 = \Delta ^{\,2}  = \left( {1 - a} \right)^{\,2}  - R^{\,2}  + a^{\,2}  \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  x_{\,T}  = \left( {1 - a} \right) \hfill \cr 
  R^{\,2}  = \left( {1 - a} \right)^{\,2}  + a^{\,2}  \hfill \cr 
  0 < y_{\,T}  = \sqrt {\left( {1 - a} \right)^{\,2}  + a^{\,2}  - \left( {1 - 2a} \right)^{\,2} }  = \sqrt {2a\left( {1 - a} \right)} \quad \left| {\;0 < a < 1} \right. \hfill \cr}  \right. \cr} 
$$
and clearly, by the triangle inequality, this $R$ is greater than the other two
which are in fact $(1-a)$ and $a$.
Then I suppose you can deal with the cases in which $A$ is external
to the ellipse.
By the way, for a smooth and closed curve, either the nearest point to a given one, and the farthest, will lay on the normal to the curve at the nearest/farthest point and passing through the given one. Told in other terms it means that a circle, centered at the given point, will be tangent to the curve at both the nearest and farthest points, besides eventually to others in between. 
A: Both the nearest and farthest points of the ellipse from the point $A=(a,0)$ have the property that the normal line to the ellipse at these points passes through $A$. The normal line through a point $(x_0,y_0)$ on the ellipse has the equation $$y_0x-(2x_0-1)y+(x_0-1)y_0 = 0$$ and plugging in the coordinates of $A$ yields the equation $(x_0+a-1)y_0=0$, so either $x_0=1-a$ or $y_0=0$.  
Considering first the condition $x_0=1-a$, the corresponding $y$-coordinates are $\pm\sqrt{2a(1-a)}$, which are nonzero for $0\lt a\lt1$, i.e., when $A$ is interior to the ellipse. The distance of these two points from $A$ is then easily found to be $d=\sqrt{2a^2-2a+1}$. For $y_0=0$, the candidate points are, of course the endpoints $(0,0)$ and $(1,0)$ of the minor axis, with respective distances $|a|$ and $|a-1|$ from $A$. For an exterior point (or a point on the ellipse), these latter two points are the only possibilities, so the farthest point on the ellipse from $A$ in that case is the further endpoint of the minor axis. On the other hand, it’s not too hard to show that $0\lt a\lt 1$ implies that both $d\gt|a|$ and $d\gt|a-1|$, so those off-axis points are the farthest from $A$ when it is interior to the ellipse. This is easy to see geometrically: if $A$ is a point on the minor axis, a circle centered on $A$ that passes through an endpoint of the axis is internally tangent to the ellipse, so every other point on the ellipse is farther from $A$ than either of these endpoints.  
In fact, a similar geometric argument shows without performing any of the above calculations that for a point on the $x$-axis external to this ellipse, the farthest point of the ellipse from it must be the far endpoint $B$ of the minor axis. Consider again the circle centered at $A$ with radius $AB$. This ellipse is internally tangent to this circle at $B$, so every other point on the ellipse must be closer to $A$ than $B$.
