Question

a problem for u
Two equations in a plane are given
a line y=2b(a-x)
a parabola y=x^2-2ax+1

Answer 1

That isn't a question or a problem.

Answer 2

\[A=\left\{ (a,b) \in R^2 \ such \ that \ \ line \ and \ Parabola \ dont \ intercept \ \right\}\]

Answer 3

find the area of A

Answer 4

intersect*

Answer 5

I don't completely understand.

Answer 6

A is a set of points (a,b) such that line and parabola dont intersect each other
problem is finding area of A

Answer 7

It's the area of the part where they don't intersect?

Answer 8

can you draw it out? And 'show' what you mean.

Answer 9

i cant draw it because a and b are unknown

Answer 10

solution would be 2ab < 1...this are the pairs of a and b

Answer 11

answer is a^2+b^2<=1

Answer 12

@arghya
\[2b(a-x)=x^2-2ax+1 \\ x^2+2(b-a)x-2ab+1=0 \]

Answer 13

Yeah then ya will take the discrimant less than zero..

Answer 14

yes thats right

Answer 15

I did the same..but did a silly mistake,so was not getting

Answer 16

\[\Delta<0 \\ (2(b-a))^2-4(-2ab+1)<0 \\ 4(a^2+b^2-4ab)+8ab-4<0 \\ 4(a^2+b^2)<4 \\ a^2+b^2<1 \\ then \ the \ area \ of \ A \ is \ \frac{\pi}{4}\]