Question

How do I graph `2x^2+sqrt3xy+y^2-10=0`?

Answer 1

The graph of `2x^2+sqrt3xy+y^2-10=0` is an ellipse with diagonal axes. The major and minor axes are not parallel with the x and y axes.

Explanation:

Solve for y in terms of x then solve for some (x, y) coordinates to draw its graph

Using quadratic formula

rewrite the equation first, so that it appears as follows

`y^2+sqrt3xy+2x^2-10=0`

Let `a=1` and `b=sqrt3x` and `c=2x^2-10`

`y=(-b+-sqrt(b^2-4ac))/(2a)`

`y=(-sqrt3x+-sqrt((sqrt3x)^2-4(1)(2x^2-10)))/(2*1)`

`y=(-sqrt3x+-sqrt(40-5x^2))/2`

Compute y values for some x arbitrary values

`y=(-sqrt3x+sqrt(40-5x^2))/2`

`" "x" " " " " " "y`
`" "0" " " " "+3.16`
`+0.5" " " " "2.679`
`+0.75" " " " "2.399`
`+1" " " " "2.092`
`+1.25" " " " "1.754`
`+1.5" " " " "1.3819`
`+1.75" " " " ".9687`
`+2" " " " "0.504`
`+2.25" " " " "-0.0323`
`+2.5" " " " "-0.68604`
`+sqrt3" " " " "1`

`" "x" " " " " " "y`
`-0.5" " " " "3.54548`
`-0.75" " " " "3.698`
`-1" " " " "3.824`
`-1.25" " " " "3.919`
`-1.5" " " " "3.9799`
`-1.75" " " " "3.9998`
`-2" " " " "3.968`
`-2.25" " " " "3.864`
`-2.5" " " " "3.644`
`-sqrt3" " " " "4`

Kindly see the graph of `y^2+sqrt3xy+2x^2-10=0`

graph{y^2+sqrt3xy+2x^2-10=0[-sqrt8,sqrt8,-5,5]}

God bless you...