Question

Suppose $x$ and $y$ are real numbers and that $x^2+9y^2-4x+6y+4=0$ , then the minimum value of $\left(4x-9y\right)$ is

• A. $1$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (D)

$x^2+9y^2-4x+6y+4=0$
$\Rightarrow {\left(x-2\right)}^2+9\left(y^2+\frac{2}{3}y+\frac{1}{9}\right)=1$
$\Rightarrow \frac{{\left(x-2\right)}^2}{1}+\frac{{\left(y+\frac{1}{3}\right)}^2}{\left(\frac{1}{9}\right)}=1$
Its parametric equation is
$x=2+\cos \theta ,y=-\frac{1}{3}+\frac{1}{3}\sin \theta$
Hence, $4x-9y=8+4\cos \theta +3-3\sin \theta$
$\Rightarrow$ Minimum value of $4x-9y$ is $6$