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.
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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`