Question

The centre of the conic represented by the equation $14x^2-4xy+11y^2-44x-58y+71=0$ is

• A. $(3,2)$
• B. $(2,3)$
• C. $(3,3)$
• D. $(2,2)$

Answer 1

The correct option is B $(2,3)$

Let $f(x,y)=14x^2-4xy+11y^2-44x-58y+71$
Partially differentiate w.r.t. $x$ and equate to zero
$\frac{\delta f}{\delta x}=28x-4y\cdot 1-44=0$
(treat $y$ as constant)
$\Rightarrow 7x-y-11=0\cdots \left(1\right)$

Again, partially differentiate w.r.t. $y$ and equate to zero
$\frac{\delta f}{\delta y}=-4x+22y-58=0$
(treat $x$ as constant)
$\Rightarrow -2x+11y-29=0\cdots \left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right),$ we get
$x=2,y=3$
$\therefore$ Centre of given conic is $(2,3)$