Find the point on the paraboloid $z = \frac{x^2}{4}+ \frac{y^2}{25}$ that is closest to the point $(3, 0, 0)$ Find the point on the paraboloid $z = \frac{x^2}{4}+ \frac{y^2}{25}$
that is closest to the point $(3, 0, 0)$
Hi all, could someone give me a hint on how to start doing the above question?

Set $P=(3,0,0)$.
The function of interest:
$$f(x,y)=\left(\frac{x}{2}\right)^2+\left(\frac{y}{5}\right)^2$$
Its graph is given by:
$$\phi (u,v)=(u,v,f(u,v))$$
We calculate the normal-vector as follows:
$$\frac{\partial \phi }{\partial u}\times \frac{\partial \phi }{\partial v}(u,v)=\left(-\frac{u}{2},-\frac{1}{25} (2 v),1\right)$$
The normal-vector is perpendicular on the surface.
We now consider the straight-line:
$$\phi (u,v)+t \left(-\frac{u}{2},-\frac{1}{25} (2 v),1\right)$$
and ask, were this line meets P. We are going to solve
$$\phi (u,v)+t \left(-\frac{u}{2},-\frac{1}{25} (2 v),1\right)=P$$
We found these values:
$$u=2,v=0,t=-1$$
There exits other values too, but they are no reals.
Because of
$$\phi (2,0)-P=(-1,0,1)$$
this shortest distance is $\sqrt{2}$.
Have a look what we've done:

Red line for x-Axis, blue line? Our shortest one.
Were they meet is point P.
