Prove that if the equation `x^2+9y^2-4x+3=0` is satisfied for real values of `xa n dy ,t h e nx` must lie between 1 and 3 and`y` must lie between-1/3 and 1/3.

Explanation

Given equation is
`x^(2) + 9y^(2) - 4x + 3 =0` (1)
or `x^(2) - 4x + 9y^(2) + 3 = 0`
Since x is ral, we have
`(-4)^(2) - 4(9y^(2)+ 3) ge 0`
or `16 - 4 (9y^(2) + 3) ge 0`
or ` 4 - 9y^(2) - 3 ge 0`
or `9y^(2) - 1 le 0`
or ` 9y^(2) le 1`
or `y^(2)le (1)/(9)`
`rArr -(1)/(3) le y le (1)/(3)` (2)
Equation (1) can also be written as
`9y^(2) + 0y + x^(2) - 4x + 3 = 0` (3)
Since y is real, so
or `0^(2) - 4x + 3 le 0` (4)
or (x - 3) (x - 1) le 0`
or ` 1 le x le 3`