JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 9)
If x2 + 9y2 $$-$$ 4x + 3 = 0, x, y $$\in$$ R, then x and y respectively lie in the intervals :
$$\left[ { - {1 \over 3},{1 \over 3}} \right]$$ and $$\left[ { - {1 \over 3},{1 \over 3}} \right]$$
$$\left[ { - {1 \over 3},{1 \over 3}} \right]$$ and [1, 3]
[1, 3] and [1, 3]
[1, 3] and $$\left[ { - {1 \over 3},{1 \over 3}} \right]$$
Explanation
x2 + 9y2 $$-$$ 4x + 3 = 0
(x2 $$-$$ 4x) + (9y2) + 3 = 0
(x2 $$-$$ 4x + 4) + (9y2) + 3 $$-$$ 4 = 0
(x $$-$$ 2)2 + (3y)2 = 1
$${{{{(x - 2)}^2}} \over {{{(1)}^2}}} + {{{y^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} = 1$$ (equation of an ellipse).
As it is equation of an ellipse, x & y can vary inside the ellipse.
So, $$x - 2 \in [ - 1,1]$$ and $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$
x $$\in$$ [1, 3] $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$
(x2 $$-$$ 4x) + (9y2) + 3 = 0
(x2 $$-$$ 4x + 4) + (9y2) + 3 $$-$$ 4 = 0
(x $$-$$ 2)2 + (3y)2 = 1
$${{{{(x - 2)}^2}} \over {{{(1)}^2}}} + {{{y^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} = 1$$ (equation of an ellipse).
As it is equation of an ellipse, x & y can vary inside the ellipse.
So, $$x - 2 \in [ - 1,1]$$ and $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$
x $$\in$$ [1, 3] $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$
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