Question

If $x^2+9y^2-4x+3=0,x,y\in R,$ then $x$ and $y$ respectively lie in the intervals

• A. $[1,3]$ and $[1,3]$
• B. $[-\frac{1}{3},\frac{1}{3}]$ and $[-\frac{1}{3},\frac{1}{3}]$
• C. $[-\frac{1}{3},\frac{1}{3}]$ and $[1,3]$
• D. $[1,3]$ and $[-\frac{1}{3},\frac{1}{3}]$

Answer 1

The correct option is D $[1,3]$ and $[-\frac{1}{3},\frac{1}{3}]$

Given: $x^2+9y^2-4x+3=0$
$\Rightarrow (x^2-4x+4)+9y^2-1=0$
$\Rightarrow (x-2{)}^2+9y^2=1$
$\Rightarrow \frac{(x-2{)}^2}{1}+\frac{y^2}{{\left(\frac{1}{3}\right)}^2}=1$
Since it is equation of an ellipse, therefore $x$ and $y$ can vary inside the ellipse.
$x-2\in [-1,1]$
$\Rightarrow x\in [1,3]$
and $y\in [-\frac{1}{3},\frac{1}{3}]$