Question

If A(a^2,2a) ,B(1/a^2,-2/a) and S(1,0),then prove that 1/SA + 1/SB =1

Answer 1

A (a²,2a) , B (1/a²,-2/a) , S (1,0)
We have to prove that,
1/SA + 1/SB = 1
Now, SA = √[(a²-1)²+(2a-0)²]
SA = √(a⁴-2a²+1+4a²)
SA = √(a⁴ + 2a² +1)
SA = √(a² +1)²
SA = a² +1
1/SA = 1 / (a² +1)
Now, SB = √[(1/a²-1)² + (-2/a - 0)²]
SB = √(1/a⁴ - 2/a² + 1 + 4/a²)
SB = √(1/a⁴ +2/a² +1)
SB = √(1/a² + 1)²
SB = 1/a² + 1
SB = (a² + 1) / a²
1/SB = a² / (a² + 1)
1/SA + 1/SB = [1 / (a² +1)] + [a² / (a² + 1)]
1/SA + 1/SB = (1 + a²) / (1 + a²)
1/SA + 1/SB = 1
Hence proved.

Answer 2

Answer:

1

Step-by-step explanation:

A (a²,2a) , B (1/a²,-2/a) , S (1,0)

We have to prove that,

1/SA + 1/SB = 1

Now, SA = √[(a²-1)²+(2a-0)²]

SA = √(a⁴-2a²+1+4a²)

SA = √(a⁴ + 2a² +1)

SA = √(a² +1)²

SA = a² +1

1/SA = 1 / (a² +1)

Now, SB = √[(1/a²-1)² + (-2/a - 0)²]

SB = √(1/a⁴ - 2/a² + 1 + 4/a²)

SB = √(1/a⁴ +2/a² +1)

SB = √(1/a² + 1)²

SB = 1/a² + 1

SB = (a² + 1) / a²

1/SB = a² / (a² + 1)

1/SA + 1/SB = [1 / (a² +1)] + [a² / (a² + 1)]

1/SA + 1/SB = (1 + a²) / (1 + a²)

1/SA + 1/SB = 1

Hence proved.