Please show work =)
The distance between two points in the xy plane is:
D = [(x2-x1)^2 + (y2-y1)^2]^0.5
Let (x1,y1) be the point (3,0), then D is:
[(x2-3)^2 + (y2)^2]^0.5, or to save clutter:
D = [(x-3)^2 + y^2]^0.5
y = 8/x, so D = [(x-3)^2 + 64x^-2]^0.5
dD/dx = 0.5 [(x-3)^2 + 64x^-2]^-0.5 * [2(x-3) – 128x^-3]
D is at the minimum when dD/dx = 0, so solve for:
[2(x-3) – 128x^-3] = 0
x = 4, so y = 2.
The hyperbola is closest to (3, 0) at (4, 2) where the distance is the square root of 5.
This is the collective voice of the talented editorial team at Answerprime.com. Committed to delivering high-quality content, this account represents the collaborative efforts of knowledgeable experts. With a passion for research and a dedication to accuracy, Answerprime brings you insightful articles across various subjects. Explore a diverse range of expertise and perspectives curated by the editorial team through the lens of Answerprime.
Latest posts by Answer Prime (see all)
- How many 1 cm squares would it take to construct a square that is 3 m on each side? - October 16, 2025
- Can I cash my paycheck at hannaford for free? - October 16, 2025
- What is the act of ‘parking a horse’ called? - October 16, 2025