Question

if coordinate of A is (a² ,2a) and B is (1/a²,-2/a ) and S is (1,0) then find 1/SA + 1/SB = a) 2 b) 1/2 c) 1 d) 1/3​

Answer 1

Answer:

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 + (-2a-0)²]

SB = √(1/a⁴-2/a² + 1)

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:

SA=(1−a2)2+(−2a)2

=1+a4−2a2+4a2

=1+a4+2a2

=(1+a2)2

=(1+a2) 

SB=(1−a21)2+(a2)2

=1+a41−a22+a24

=1+a41+a22