Find the shortest distance from the origin to the hyperbola $x^2+8xy+7y^2=225$ 
Find the shortest distance from the origin to the hyperbola $x^2+8xy+7y^2=225$

i know that
$$d(x_0, x) = \sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$$
I also found this formula in my notes
$$ d(x_0,p) = \frac{|ax_0+by_0+cz_{0}-c|}{\sqrt{a^2+b^2+c^2}}$$
I just haven't seen it been applied in class so i'm a bit confused.
This was in a week we were learning about Lagrange Multipliers and we don't seem to be given a constraint.
 A: let $P(rcost,rsint)$ be a Point on Hyperbola. so its distance from $(0,0)$ is $r$, so we need to find Minimum value of $r$. Since $P$ lies on Hyperbola 
$$r^2cos^2t+8r^2sintcost+7r^2sin^2t=225$$ $\implies$
$$r^2=\frac{450}{8sin2t-6cos2t+8}$$
Now max value of $$8sin2t-6cos2t+8$$ is $$\sqrt{8^2+6^2}+8=18$$
hence Min value of $r^2$ is $\frac{450}{18}=25$
So shortest distance is $5$
A: Using Lagrange Multipliers we get two equations 
$$ (\lambda-1)x+4\lambda y=0-----(i)$$
$$4\lambda x+(7\lambda -1 )y=0-----(ii).$$
Since $(x,y)\neq(0,0)$  (does not satify given hyperbola), solving for $\lambda$,
$ \begin{vmatrix}
 (\lambda-1)& 4\lambda\\
4\lambda& (7\lambda-1)
 \end{vmatrix}$ =  $0$.
Solving above equation we get $\lambda=\frac{1}{9}, -1$.
Case (i): If $\lambda=-1$, substituting in $(i)$   gives $x=-2y$. Substituting in hyperbola we get $-5y^2=225$ which has no real solution.
Case (ii): If $\lambda=\frac{1}{9}$, substituting in $(i)$ gives $y=2x$. Substituting in hyperbola we get $45x^2=225$. So that $x^2=5$ and since $y^2=4x^2$ we get $y^2=20$.
Hence $x^2+y^2=25$. Thus required shortest distance is 5.
A: taking $d(x+y)=\sqrt{x^2+y^2}$
$$f(x,y) = x^2+y^2 +\lambda(x^2+8xy+7y^2-225)$$
$$f_x = 2x+2\lambda x + 8\lambda y =0$$
$$f_y=2y+8\lambda x + 14 \lambda y = 0$$
So i gotta solve the tree equations 
$$x^2+8xy+7y^2-225=0
$$
$$f_x = 2x+2\lambda x + 8\lambda y =0$$
$$f_y=2y+8\lambda x + 14 \lambda y = 0$$
So i need to equate for x and y by eliminating $\lambda$
i said $$x = \frac{-\lambda y}{2+2 \lambda }$$
$$y = \frac{-\lambda x}{2+14 \lambda }$$ 
From $f_x$ and $f_y$
 $$\lambda = \frac{-2x}{2x+8y}$$ 
 $$ \lambda = \frac{-2y}{8x+14y}$$ 
So solve...
$$ \frac{-2x}{2x+8y} = \ \frac{-2y}{8x+14y}$$
$$0=8y^2-8x^2-12yx$$
the remaining $\lambda x$ really complicates things
A: To minimise the distance, the gradient at that point must be tangent to the circle centred at the origin (*):
$$2x + 8 \left(y + x \frac{dy}{dx} \right) + 14y \frac{dy}{dx} = 0$$
$$\implies (8x + 14y) \frac{dy}{dx} = -(2x+8y)$$
$$\implies \frac{dy}{dx} \frac{y}{x} = -1 \implies -\frac{x+4y}{4x+7y} \frac{y}{x} = -1  \tag{*}$$
$$\implies xy + 4y^2 = 4x^2 + 7xy$$
$$\implies 2(2x^2 + 3xy - 2y^2) = 0 \implies (2x - y)(x + 2y) = 0 \implies y = 2x,-\frac{x}{2}$$
and factorising the condition, $\left(x+y\right)\left(x+7y\right)=225$, thus $(3x)(15x) = 225 \Rightarrow x = ±\sqrt{5}$, $y = ±2 \sqrt{5}$, or $\left(\frac{1}{2} x\right)\left(-\frac{5}{2} x \right) =225$ which is impossible since the LHS is always not positive.
Hence the minimum distance is $\sqrt{x^2+y^2} = \sqrt{5 + 4 \cdot 5} = \boxed{5}$.
