Finding point(s) on ellipse closest to origin The question is asking for the point(s)that are the closest to the origin on the ellipse 
$5x^2-6xy+5y^2=8$
I used the distance formula $d^2=x^2+y^2$
And found the x and y components to be
$2x=(10x-6y) \lambda$ and $2y=(10y-6x) \lambda$.
$\frac{2x}{2y} =\frac{(10x-6y)\lambda}{(10y-6x)\lambda}$
$10xy-6x^2 =10xy-6y^2$
$-6x^2=-6y^2$
$x^2=y^2$
$x=y$ 
$5x^2 -6x^2+5x^2 =8$ => $4x^2 =8$ => $x=\pm2$ ; $y=\pm2$ 
When I solved for x and y I got both of them to be 
$x=\pm 2$
 and $y=\pm2$
I am not sure if this is correct not but when I graphed the ellipse my points appear to be outside of the ellipse. 
Please let me know if this is the correct method to solve this and if I am doing it correctly. Thank You
 Image of ellipse
 A: Take the derivative of
$$5x^2-6xy+5y^2=8$$
$$y’=\frac{3y-5x}{5y-3x}$$
If the point on the curve is the closest to the origin, the following relationship can be established,
$$ -\frac xy =\frac{3y-5x}{5y-3x}$$
which leads to 
$$x=\pm y$$
Plug $x=-y$ back into the original equation to get
$$x^2=\frac 12$$
Thus, the closest points are $(\frac{1}{\sqrt 2},-\frac{1}{\sqrt 2})$ and $(-\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$. 
Note that the line connecting these two points is the minor axis. The other solution $x=y$ leads to the two points along the major axis, which are not the closest. You were focusing on $x=y$ instead of $x=-y$.
A: You were doing well until you went from $x^2=y^2$ to $x=y$: another possibility is $x=-y$. The second mistake came at $4x^2=8$: the solutions are $\pm\sqrt2$, not $\pm2$, which is why the points weren’t on the ellipse. This gives you one pair of points, $\pm(\sqrt2,\sqrt2)$, at a distance of $2$ from the origin.  
Now you have to try the other possible solution to $x^2=y^2$. Setting $y=-x$, we eventually get the equation $16x^2=8$, which yields the points $\pm(1/\sqrt2,-1/\sqrt2)$, at a distance of $1/2$ from the origin, so these are the solutions to your exercise.
