If x2 + 9y2 $-$ 4x + 3 = 0, x, y $\in$ R, then x and y respectively lie in the intervals :
Solution
x<sup>2</sup> + 9y<sup>2</sup> $-$ 4x + 3 = 0<br><br>(x<sup>2</sup> $-$ 4x) + (9y<sup>2</sup>) + 3 = 0<br><br>(x<sup>2</sup> $-$ 4x + 4) + (9y<sup>2</sup>) + 3 $-$ 4 = 0<br><br>(x $-$ 2)<sup>2</sup> + (3y)<sup>2</sup> = 1<br><br>$${{{{(x - 2)}^2}} \over {{{(1)}^2}}} + {{{y^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} = 1$$ (equation of an ellipse).<br><br>As it is equation of an ellipse, x & y can vary inside the ellipse.<br><br>So, $x - 2 \in [ - 1,1]$ and $y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$<br><br>x $\in$ [1, 3] $y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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