Find the parameter m accordingly : $$ X/5 + Y/4 + Z/m = 1 $$ is the tangent plane of the ellipsoid $$ (X^2)/9 + (Y^2)/4 + (Z^2)/1 = 1 $$ 
So I presume the tangent plane of the ellipsoid shape is in M(Xo,Yo,Zo).
That means:
$$ F(x,y,z)= (X^2)/9 + (Y^2)/4 + (Z^2)/1 -1 = 0 $$;
$$ \delta F / \delta X = (2/9)*X $$
$$ \delta F / \delta Y = (1/2)*Y $$
$$ \delta F/ \delta Z= 2*Z $$
Which means:
$$ \delta F / \delta X (M) = (2/9)*Xo $$
$$ \delta F / \delta Y (M) = (1/2)*Yo $$
$$ \delta F/ \delta Z (M)= 2*Zo $$
SO the tangent plane is:
$$ \Sigma_t = \delta F / \delta X (M) *(X-Xo) + \delta F / \delta Y (M)*(Y-Yo) + \delta F/ \delta Z (M)*(Z-Zo) $$ 
$$ \Sigma_t = (2/9)*Xo*(X-Xo)+ (1/2)*Yo*(Y-Yo)+ 2*Zo*(Z-Zo) $$
$$ \Sigma_t = (2/9)*Xo*X + (1/2)*Yo*Y + 2*Zo*Z = (2/9)*(Xo)^2 + (1/2)*(Yo)^2 + 2*(Zo)^2 $$
I don't know how to proceed.
Please explain every step of the process if you can.
 A: Any point on the ellipsoid can be written as $(3\cos\theta\cos\phi, 2\cos\theta\sin\phi, \sin\theta)$. The normal to the ellipsoid at this point is $$\left(2X/9, 2Y/4, 2Z\right)$$ and hence the tangent plane is 
$$(X-3\cos\theta\cos\phi)(\cos\theta\cos\phi/3) + (Y-2\cos\theta\sin\phi)(\cos\theta\sin\phi/2)+(Z-\sin\theta)\sin\theta = 0$$
If this is the given plane, then comparing the coefficients and the constant terms, we have
$$5\cos\theta\cos\phi/3 = 4\cos\theta\sin\phi/2 = m\sin\theta = 1 $$
and hence
$$9/25 + 1/4 + 1/m^2 = 1$$ and hence $m = \pm 10/7$
There is a shorter way as well:
Consider the mapping $F: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $F(X,Y,Z) = (3X, 2Y, Z)$. This maps the ellipsoid to the sphere $X^2+Y^2+Z^2 = 1$ and the plane to $3X/5+2Y/4 + Z/m = 1$. The condtion translates to the plane being a tangent plane to the sphere. Thus the perpendicular distance from the center of the sphere to the plane equals the radius of the sphere. Thus 
$$\left|\frac{1}{\sqrt{\frac{9}{25}+\frac{1}{4}+\frac{1}{m^2}}}\right| = 1$$ and we obtain values of $m$ as before.
