Given $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1$.
Then minimum value of $x^2 + y^2 = ?$
P.S. My solution: Suppose that $x = 4\cos{\theta}+12$and $y = 5\sin{\theta}-5$
and expand $x^2 + y^2$ to find minimum value, but stuck in the end.
Thank you for every comment.
HINT: use the Lagrange Multiplier Method $$F(x,y,\lambda)=x^2+y^2+\lambda((x-12)^2/16+(y+5)^2/25-1)$$ solve the system $$2\,x+\lambda\, \left( x/8-3/2 \right) =0$$ $$2\,y+\lambda\, \left( {\frac {2}{25}}\,y+2/5 \right)=0 $$ $$1/16\, \left( x-12 \right) ^{2}+1/25\, \left( y+5 \right) ^{2}-1=0$$
One could find points on the ellipse where the normal line goes through the origin. There are two such points, one closest to the origin, the other furthest away from the origin. None is easily solvable as it appears.
If one uses the trig parameterization by the OP, $x = 4\cos{\theta}+12$ and $y = 5\sin{\theta}-5$ then $\frac{dx}{d\theta}=\dot x=-4\sin\theta$ and $\frac{dy}{d\theta}=\dot y=5\cos\theta$. The vector $(\dot x,\dot y)$ is tangent, and the vector $(\dot y,-\dot x)$ is normal to the ellipse. The normal line must go through the origin hence we obtain equations $4\cos{\theta}+12+5t\cos\theta=0$ and $5\sin{\theta}-5+4t\sin\theta=0$. Hence $\displaystyle t=-\frac{4\cos{\theta}+12}{5\cos\theta}=\frac{5(1-\sin{\theta})}{4\sin\theta}$, or equivalently $-16\sin\theta(\cos\theta+3)=25\cos\theta(1-\sin{\theta})$. WA only finds approximate solutions $\theta\approx2\cdot(-0.27735)=0.5547$ and $\theta\approx2\cdot(1.3616)=2.7232$. The former produces $x\approx15.40023394,y\approx-7.63344153181$ (furthest point), the latter gives $x\approx8.34502730872,y\approx-2.96853861638$ (closest point).
Something similar is obtained if one follows the suggestion by OP to plug $x = 4\cos{\theta}+12$ and $y = 5\sin{\theta}-5$ in $r^2=x^2+y^2$ and find its minimum. We have $r^2=16\cos^2\theta+96\cos\theta+25\sin^2\theta-50\sin\theta+169=185+96\cos\theta+9\sin^2\theta-50\sin\theta$ and $\displaystyle \frac{dr^2}{d\theta}= 18\cos\theta\sin\theta-50\cos\theta-96\sin\theta=0$. WA gives the same solutions as above, $\theta\approx2\cdot(-0.27735)=0.5547$ and $\theta\approx2\cdot(1.3616)=2.7232$.
It could be noted that the above two methods are essentially the same. Indeed the equation $-16\sin\theta(\cos\theta+3)=25\cos\theta(1-\sin{\theta})$ could be rewritten as $9\cos\theta\sin\theta-25\cos\theta-48\sin\theta=0$ which is clearly equivalent to $18\cos\theta\sin\theta-50\cos\theta-96\sin\theta=0$.
Alternatively one may find the gradient of the ellipse to be $\displaystyle (\frac{x-12}8,\frac{2(y+5)}{25})$. Since the normal line at $(x,y)$ must meet the origin $(0,0)$ we obtain $\displaystyle x+t\frac{x-12}8=0$ and $\displaystyle y+t\frac{2(y+5)}{25}=0$, hence $\displaystyle t=\frac{-8x}{x-12}=\frac{-25y}{2(y+5)}$. Thus $16x(y+5)=25y(x-12)$. Solving the latter simultaneously with the equation of the ellipse gives, using WA, $x\approx8.34504, y\approx-2.96851$ (closest point), and $x\approx15.4002, y\approx-7.63343$ (furthest point).
Solving $16x(y+5)=25y(x-12)$ for $y$ we obtain $\displaystyle y=\frac{80x}{3(3x-100)}$. Substituting in the equation of the ellipse, we have $\displaystyle \frac{ 81x^4-7344x^3+239968x^2-3091200x+12960000}{144(9x^2-600x+10000)}=0$. Since the numerator is a quartic, it is possible to find exact solutions (but it would be tedious and not easily done, and may not necessarily help understand the solution better). Here are one more time the approximate solutions by WA.
This third method is essentially the same as using Lagrange multipliers (suggested in the answer by Dr. Sonnhard Graubner). Excluding $t$ from $\displaystyle x+t\frac{x-12}8=0$ and $\displaystyle y+t\frac{2(y+5)}{25}=0$ results in $\displaystyle t=\frac{-8x}{x-12}=\frac{-25y}{2(y+5)}$, whereas excluding $\lambda$ from $\displaystyle 2\,x+\lambda\, \left( x/8-3/2 \right) =0$ and $\displaystyle 2\,y+\lambda\, \left( {\frac {2}{25}}\,y+2/5 \right)=0$ results in $\displaystyle \lambda=\frac{-16x}{x-12}=\frac{-50y}{2(y+5)}$,
Another, maybe not as elegant, way is to solve for $y$, getting $$y=-5\pm\frac54\sqrt{16-\left( x -12\right)^2}$$ and then, since $$y_+(x):=\left(-5+\frac54\sqrt{16-\left( x -12\right)^2}\right)^2\le\left(-5-\frac54\sqrt{16-\left( x -12\right)^2}\right)^2=:y_-(x), $$ find the zero $x_0$ of the derivative of $x^2+(y_+(x))^2$ , which is $$-\frac98 x +\frac{25}{2}\left(3+\frac{x-12}{\sqrt{16-(x-12)^2}}\right) \tag{$\star$}$$ and compute $x_0^2+(y_+(x_0))^2$. Note that $x_0$ is unique and is approximately $8.345$. The other solution mentioned by Mirko is produced by squaring, which is necessary to solve exactly $(\star)$, ending up with $$\frac{(x-12)^2}{16-(x-12)^2}=\left(\frac{9}{100}x-3\right)^2,$$ whose solutions are indeed the roots of the quartic found by Mirko.
