anonymous
  • anonymous
Find the point in the first quadrant that lies on the hyperbola y^2-x^2=5 and is closest to point (1,0). If we write the point as (a,b), then a=___ & b=___
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
a is not 0-3
anonymous
  • anonymous
if you find a, i can get b
amistre64
  • amistre64
the slope at any point is defined as x/y i can tell that much so far

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amistre64
  • amistre64
which means the slope between our given point and the nearest point on the hyper is -y/x
amistre64
  • amistre64
if im thinking this right, then x=1/2
amistre64
  • amistre64
\[(\frac{1}{2},\frac{\sqrt{19}}{2})\] if my idea is correct
anonymous
  • anonymous
we could also try this maybe would work. solve for y and get \[y^2=x^2+5\] \[y=\sqrt{x^2+5}\] and the square of the distance between \[(x, \sqrt{x^2+5}\] and \[(1,0)\] is \[(x-1)^2+\sqrt{x^2+5}^2=(x-1)^2+x^2+5\]
amistre64
  • amistre64
y^2-x^2=5 2y y' -2x = 0 y' = x/y slope from 1,0 to x,y 1 ,0 -x-y ----- 1-x,-y -y -y ---- = ---- 1-x x 1-x=x 1=2x 1/2 = x
anonymous
  • anonymous
\[D^2=2x^2-2x+6\] minimum of quadratic at \[-\frac{b}{2a}=\frac{1}{2}\] so i guess we get the same answer and as usual more than one way to skin a cat
anonymous
  • anonymous
im a little confused by how you got the 1/2 but thanks

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