Find the distance from the point $P = (9, 4, 5)$ to the ellipsoid $x^2 + 2y^2 + 4z^2 = 16$. Then, find the distance if the ellipsoid is restricted to first octant.
I tried using the distance formula according to optimization: $$d = \frac{|ax_0 + by_0 + cz_0|}{\|n\|}$$
I plugged in the coefficients (n) of the ellipsoid (1,2,4) into the formula as a,b,and c respectfully. Then, I plugged the points $9,4,5$ as $x_0$, $y_0$ and $z_0$. I also calculated the norm of $n$, $$d = \frac{|1(9) + 2(4) + 4(5)|}{\|\sqrt{21}\|}$$
I didn't really understand the other part of the first octant.
It gives me an answer but I am not sure if is the right approach. Thanks.
The normal vector to the ellipsoid at the point on the surface $Q=(x,y,z)$ is
$$n_Q=(2x,4y,8z)$$
and the condition for the distance from $P = (9, 4, 5)$ is that $PQ\parallel n_Q$ that is
$$(2x,4y,8z)=t (x-9, y-4, z-5)$$
that is $(x,y,z)=\left(\frac{9t}{t-2},\frac{4t}{t-4},\frac{5t}{t-8}\right) $ which leads numerically to the following solutions
$t\approx -1.27283 \implies Q_1\approx\left(3.500169,0.965575,0.686361\right) \implies d_1\approx 7.619954$
$t\approx 0.593020 \implies Q_2\approx\left(-3.793350,-0.696240,-0.400311\right)\implies d_2\approx 14.659053$
As an alternative by Lagrange multipliers we need to minimize
$$d(x,y,z)=(x-9)^2+(y-4)^2+(z-5)^2$$
with the condition $x^2+2y^2+4z^2-16=0$ which lead to
which lead to $(x,y,z)=\left(\frac 9{1-\lambda},\frac 2{1-2\lambda},\frac 5{1-4\lambda}\right)$ and numerically to the following solutions
$\lambda \approx -1.571305 \implies Q_1\approx\left(3.500169,0.965575,0.686361\right) \implies d_1\approx 7.619954$
$\lambda\approx 3.372573 \implies Q_2\approx\left(-3.793350,-0.696240,-0.400311\right)\implies d_2\approx 14.659053$
In both case we conclude that the distance is $d=d_1=|PQ_1|\approx 7.619954$ with $Q_1$ in the first octanct which seems an obvious result since $P$ itself is located in the first octant.
As you noticed, this part of the problem looks not very well stated.
Graphical sketch
Refer also to the related