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

Question

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.

Answer 1

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

Answer 2

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 $5$

Answer 3

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 $\sqrt{x^2+y^2} = \sqrt{5 + 4 \cdot 5} = \boxed{5}$.