Intersection Point

Algebra Level 2

If ( x , y ) (x,y) are the coordinates at which the graphs of the equations of x = 1 y x=\frac { 1 }{ y } and x 2 − y 2 = 1 { x }^{ 2 }-{ y }^{ 2 }=1 intersect in the first quadrant (both x x and y y must be positive), what is the value of x + y x+y ? Round to two decimal points. Don't use a graphing calculator.


The answer is 2.06.

Louis Ullman by Louis Ullman , 16, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
May 5, 2021

Let y = 1 x y = \frac{1}{x} and substitute it into the latter hyperbola equation:

x 2 − ( 1 x ) 2 = 1 ⇒ x 4 − x 2 − 1 = 0 ⇒ x 2 = 1 ± 1 − 4 ( 1 ) ( − 1 ) 2 ⇒ x 2 = 1 + 5 2 ⇒ x = 1 + 5 2 \large x^2 - (\frac{1}{x})^2 =1 \Rightarrow x^4 - x^2 - 1 = 0 \Rightarrow x^2 = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} \Rightarrow x^2 = \frac{1+\sqrt{5}}{2} \Rightarrow x = \sqrt{\frac{1+\sqrt{5}}{2}}

and y = 2 1 + 5 ⋅ 5 − 1 5 − 1 = 2 ( 5 − 1 ) 5 − 1 = 5 − 1 2 \large y =\sqrt{\frac{2}{1+\sqrt{5}}} \cdot \sqrt{\frac{\sqrt{5}-1}{\sqrt{5}-1}} = \sqrt{\frac{2(\sqrt{5}-1)}{5-1}} = \sqrt{\frac{\sqrt{5}-1}{2}} . Hence, x + y = 1 + 5 2 + 5 − 1 2 ≈ 2.06 . x+y = \sqrt{\frac{1+\sqrt{5}}{2}} + \sqrt{\frac{\sqrt{5}-1}{2}} \approx \boxed{2.06}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...