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I am assuming the last equation is:
x^2 + y^2 < 17
Since the hyperbolas become closer and closer the further out they get, the least possible distance would be the intersections on the circle. There are 2 intersections in each quadrant. However, we can assume the distances between all intersections are equal because of symmetry. Using the first quadrant and a TI graphing calculator to determine the intersections, the intersections are:
(3,sqrt8) and (sqrt8,3)
Using the distance formula to find the distance, we get:
d = sqrt((3 - sqrt8)^2 + (sqrt8 - 3)^2) = 0.24 units
Hi, thank you very much but I still have some question. If we don't know the gesture of the graph, can we show by reasoning from equations that the points of least possible distance are on the intersection.
The circle with radius root(17) and a hyperbola will have 1 intersection in each quadrant.
By that logic, the circle has 2 intersections in each quadrant. We can compare a point from one quadrant to another, and the distance will be much greater than when we compare the intersections inside one quadrant.
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