Answer to Question #150264 in Analytic Geometry for Asfand Khan

Question

The circle whose center is located in the first coordinate quarter touches the axis Ox at the point M, intersects two hyperbolas y=k1/x and y=k2/x (k1,k2 > 0) at the points A and B such that the line AB passes through the origin O. It is known that (4/k1) + (1/k2) = 20. Find the smallest possible length of the OM segment. In response, write down the square of the length of the segment OM.

Answer 1

$|OM|^2=|OA|\cdot|OB|$

Points $A(x_A,y_A)$ and $B(x_B,y_B)$ are points of intersection circle with hyperbolas.

$|OM|^2=\sqrt{(x_A^2+y_A^2)(x_B^2+y_B^2)}$

$|OM|^2=\sqrt{(x_A^2+(k_1/x_A)^2)(x_B^2+(k_2/x_B)^2)}$

Since A and B lie on the same line, then: $x_A/y_A=x_B/y_B$

$\frac{x_Ay_A}{x_By_B}=\frac{k_1}{k_2}\implies (\frac{x_A}{x_B})^2=\frac{k_1}{k_2}$

$|OM|^2=\sqrt{\frac{(x_A^4+k_1^2)(x_B^4+k_2^2)}{x_A^2x_B^2}}=\sqrt{x_A^2x_B^2+2k_1k_2+k_1^2k_2^2/x_A^2x_B^2}$

$x_B^2=\frac{k_2}{k_1}x_A^2$

$|OM|^2==\sqrt{\frac{k_2}{k_1}x_A^4+2k_1k_2+k_1^3k_2/x_A^4}$

$(4/k1) + (1/k2) = 20$

$k_2=\frac{k_1}{20k_1-4}$

$|OM|^2==\sqrt{\frac{k_1x_A^8+2x_A^4k_1^3+k_1^5}{x_A^4k_1(20k_1-4)}}$

$|OM|^2=\frac{1}{x_A^2}\sqrt{\frac{x_A^8+2x_A^4k_1^2+k_1^4}{20k_1-4}}$

Let: $x=x_A, k=k_1, |OM|=y(x)$

Then:

$y(x)=\frac{1}{x}(\frac{x^8+2x^4k^2+k^4}{20k-4})^{1/4}$

$y'(x)=\frac{1}{4}(\frac{x^4+2k^2}{20k-4}+\frac{k^4}{x^4(20k-4)})^{-3/4}\cdot(\frac{4x^3}{20k-4}-\frac{4k^4}{x^5(20k-4)})=0$

$4x^8-4k^4=0$

$x=\sqrt{k}$

$x_A=\sqrt{k_1}$

$|OM|_{min}^2=\frac{1}{k_1}\sqrt{\frac{k_1^4+2k_1^4+k_1^4}{20k_1-4}}=\frac{k_1}{\sqrt{5k_1-1}}$