Question

The centre of the conic section $2x^2+4y^2+2xy+4x-6y+17=0$ is

• A. $(\frac{11}{7},-\frac{8}{7})$
• B. $(\frac{11}{7},\frac{8}{7})$
• C. $(-\frac{11}{7},\frac{8}{7})$
• D. $(-\frac{11}{8},\frac{7}{8})$

Answer 1

The correct option is C $(-\frac{11}{7},\frac{8}{7})$

Let $f(x,y)=2x^2+4y^2+2xy+4x-6y+17=0$
Partially differentiate w.r.t. $x$
$\frac{\delta f}{\delta x}=4x+0+2y+4-0+0=0$ (treat $y$ as constant)
$\Rightarrow 2x+y+2=0\cdots \left(i\right)$

Again, partially differentiate w.r.t. $y$
$\frac{\delta f}{\delta y}=0+8y+2x-6+0=0$ (treat $x$ as constant)
$\Rightarrow x+4y-3=0\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$
$x=-\frac{11}{7}$ and $y=\frac{8}{7}$
Then centre is $(-\frac{11}{7},\frac{8}{7})$