Find the point on the graph of y Squareroot x2 1 that is c

Find the point on the graph of y = Squareroot x^2 - 1 that is closest to the point (0, 1). (1.12, 0.50)

Solution

point on graph y=(x²-1) is (x,y)

distance detween point on graph (x,y) and (0,1) is d=[x²+(y-1)²]

d=[x²+((x²-1) -1)²]

d=[x²+(x²-1) +1-2(x²-1)]

d=[2x²-2(x²-1)]

(x,y) is closest point to (0,1) when distance is minimum

distance is minimum when f(x)=2x²-2(x²-1) is minimum

f(x) is ,minimum when f \'(x)=0

f \'(x)=4x -(2*2x/_2(x²-1)) =0

f \'(x)=4x -(2x/_(x²-1)) =0

2x[2-(1/_(x²-1))]=0

x=0 ,2-(1/_(x²-1))=0

2= 1/_(x²-1)

=>2(x²-1) =1

=>4(x²-1)=1

=>4x²-4=1

=>4x²=5

=>x=(5)/2 .x=-(5)/2

x=0 ,=>y is undefined

x=-(5)/2 => y=((-(5)/2)²-1) =1/2

x=(5)/2 => y=(((5)/2)²-1) =1/2

x=-(5)/2,d=[2x²-2(x²-1)]

d=[2(-(5)/2)²-2((-(5)/2)²-1)] =(3/2)

x=(5)/2,d=[2x²-2(x²-1)]

d=[2((5)/2)²-2(((5)/2)²-1)] =(3/2)

coordinates of closest points are (-(5)/2 ,1/2) ,((5)/2 ,1/2)

coordinates of closest points are (-1.12 ,0.5) ,(1.12 ,0.5)