Question

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=___

Answer 1

a is not 0-3

Answer 2

if you find a, i can get b

Answer 3

the slope at any point is defined as x/y i can tell that much so far

Answer 4

which means the slope between our given point and the nearest point on the hyper is -y/x

Answer 5

if im thinking this right, then x=1/2

Answer 6

\[(\frac{1}{2},\frac{\sqrt{19}}{2})\]
if my idea is correct

Answer 7

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
\[(x-1)^2+\sqrt{x^2+5}^2=(x-1)^2+x^2+5\]

Answer 8

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

Answer 9

\[D^2=2x^2-2x+6\] 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

Answer 10

im a little confused by how you got the 1/2 but thanks