Question

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