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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. |
Answer» 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` |
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