Question

The points on the curve x^2-y^2=4 closest to the point (6 0) are

Answer 1

let the point on the curve is (w,sqrt(w^2-4)) 

d^2=x^2+y^2 

d^2=(w-6)^2+w^2-4 

=2w^2-12w+32 

2dd'=4w-12=0 for crit 

w=3 

d"=4<0 and so it is minimum distance 

the point is (3,+-sqrt5)

HOPE IT HELPS YOU

☺☺☺☺☺☺☺

Answer 2

Answer:

The points on the curve x^2-y^2=4 closest to the point (6 0) are

x^2-y^2=4

We know that distance between two points is given by d =√(x-a) 2 +(y-b) 2

Here a=6 b=0

On squaring both sides we get

d2=(x-a) 2 +(y-b) 2

=(x-6) 2 +(y-0) 2

=(x-6) 2 +(y) 2

But x^2-y^2=4

So y^2=x^2- 4

=(x-6) 2 +(y) 2

=(x-6) 2 +x^2- 4

f(x) =(x-6) 2 +x^2- 4

On differentiating both sides we get

f'(x) = 2(x-6) +2x

On equating to 0

2(x-6) +2x =0

2x-12 +2x=0

4x -12=0

4x =12

x=3

y^2=x^2- 4

y^2 = 9-4

y^2= 5

y =√5

(x, y) = (3, √5)