# 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.

• Distance to origin $\sqrt{x^2+y^2}$. Constraint: your hyperbola's formula. – A.S. Oct 21 '15 at 1:23
• Contestant equals $x^2+8xy+7y^2-225=0$ – kingportable Oct 21 '15 at 1:25
• $f(x,y) = \sqrt{x^2+y^2}-\lambda(x^2+8xy+7y^2-225)$ and find the partial derivatives? – kingportable Oct 21 '15 at 1:55
• much better to use the squared distance, $x^2 + y^2.$ The $(x,y)$ location(s) where that is smallest is(are) also the places where the distance from the origin is smallest. – Will Jagy Oct 21 '15 at 1:59
• Hey guys i got a bit further could you see what i did? – kingportable Oct 21 '15 at 2:45

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 $5$

• Hi Ekaveera thanks a lot, but i think i'm suppose to use Lagrange multipliers. – kingportable Oct 21 '15 at 5:14

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

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.