Shortest Distance Problem

Let us take an arbitrary line through the origin $y = mx$ that intersects the above hyperbola in the points;

$x^2 + 8x(mx) + 7(mx)^2 = 225 \Rightarrow (1 + 8m + 7m^2)x^2 = 225 \Rightarrow x = \pm \frac{15}{\sqrt{1 + 8m + 7m^2}}$ and $y = \pm \frac{15m}{\sqrt{1 + 8m + 7m^2}}$ .

Let $L(m) = \sqrt{x^2 + y^2} = \sqrt{(\frac{15}{\sqrt{1 + 8m + 7m^2}})^2 + (\frac{15m}{\sqrt{1 + 8m + 7m^2}})^2} = 15 \cdot \sqrt{\frac{1 + m^2}{1 + 8m + 7m^2}}$ ;

be the distance from the origin to either of these points and taking $\frac{dL}{dm} = 0$ gives:

$\frac{dL}{dm} = \frac{15}{2} \cdot \frac{\sqrt{1 + 8m + 7m^2}}{\sqrt{1 + m^2}} \cdot \frac{(2m)(1+8m+7m^2) - (1+m^2)(8+14m)}{(1+8m+7m^2)^2} = 0 \Rightarrow m = 2, -\frac{1}{2}.$

Upon observation, the hyperbola has negative slopes in both of its branches. The shortest distance from the origin to the hyperbola will occur along a line that is normal to the hyperbola (i.e. the product of $m$ and the slope of the hyperbola's tangent must equal -1). This in turn forces the critical value of $m$ to be positive (discounting $m = -\frac{1}{2}$ ).

Hence, the shortest distance from the origin to the hyperbola occurs along the line $y = 2x$ and computes to $L(2) = \boxed{5}.$

If you want to find the actual answer, that is, avoid using any approximations, you can use a Lagrange multiplier, minimizing distance subject to the constraint of the given equation. The answer comes out to be 5.

If y=0 then x^2-225=0 then x=15 If x=0 then y=15/sqrt(7) Then the shortest distance be 5

put x=0 ---> 7y^2 = 225 ---> (take absolute value) y=15/sqrt(7) put y=0 ---> x^2=225 ---> (take absolute walue) x=15

the shotest distance is perpendicular to the hyperbola from the origin using T rule (method) xy = shortest distance * hypotenuse (15*15)/sqrt(7) = shortest distance * sqrt(15^2 + 225/7) solving ....shortest distance = 5 (by approximation)

THANKS! solved by (yury)