Question

The centre of the ellipse 14x2 - 4xy + 11y2 - 44x - 58y + 71 = 0 is

Answer 1

Useful fact : For conic f(x,y) = 0 centre is found by solving ∂f/∂x = ∂f/∂y = 0

f(x,y) = 14x²−4xy+11y²−44x−58y+71

∂f/∂x = 28x−4y−44 = 0 → 7x−y = 11

∂f/∂y = −4x+22y−58 = 0 → −2x+11y = 29

These solve to give centre as (2,3)

Answer 2

The center of the ellipse is (2,3).

Step-by-step explanation:

The given equation of ellipse is

$14x^2-4xy+11y^2-44x-58y+71$

If equation if ellipse is $F(x,y)=ax^2+2bxy+cy^2+2dx+2ey+2f=0$, then we know to solve two equations $\frac{\partial F}{\partial x}=0$ and $\frac{\partial F}{\partial y}=0$.

For given equation

$\frac{\partial F}{\partial x}=28x-4y-44=0$

$28x-4y=44$                 .... (1)

$\frac{\partial F}{\partial y}=-4x+22y-58=0$

$-4x+22y=58$           .... (2)

Multiply both sides by 7.

$-28x+154y=406$            .... (3)

Add (1) and (3).

$-28x+154y+28x-4y=406+44$

$150y=450$

$y=3$

The value of y is 3.

Substitute y=3 in equation (1).

$28x-4(3)=44$

$28x-12=44$

$28x=44+12$

$28x=56$

$x=2$

Therefore, the center of the ellipse is (2,3).

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