Question

1) Find distance AB if x= 4, y = -5​

Answer 1

Answer:

ANSWER

Given curve is y=

x

.... (i)

Since, the point (a,b) is on the curve (i).

Then, b=

a

⇒b

2

=a .... (ii)

Let D= distance between (a,b) and (1,0).

∴D

2

=(a−1)

2

+(b−0)

2

=a

2

+1−2a+b

2

⇒D

2

=a

2

+1−2a+a [from Eq. (ii)]

⇒D

2

=a

2

−a+1 .... (iii)

Let P=a

2

−a+1

Now, P is max/min according as D

2

is max/min.

∴

da

dP

=2a−1

and

da

2

d

2

P

=2>0 (min)

For maximum or minimum of P,

put

da

dP

=0⇒2a−1=0

⇒a=

2

1

At (a=

2

1

), point (a,b) is closest to the point (1,0)

From Eq. (ii), we get

b

2

=

2

1

⇒b=

2

1

Therefore, ab=

2

1

⋅

2

1

=

2

2

1

=

4

2