The equation of the tangent line to the curve at the given point is
y = -x = 2
The formula for calculating the equation of a line in point-slope form is expressed as:
[tex]y-y=m(x-x_0)[/tex] where:
m is the slope of the line
(x0, y0) is the point on the line
Given the equation [tex]2x^2+xy+2y^2=5[/tex]
Differentiate the given equation implicitly to get the slope
[tex]4x+x\frac{dy}{dx}+y+4y\frac{dy}{dx} = 0\\4x+y=-x \frac{dy}{dx}-4y\frac{dy}{dx}\\\frac{dy}{dx}=\frac{4x+y}{-x-4y} \\\frac{dy}{dx}=\frac{4+1}{-1-4}\\\frac{dy}{dx}=\frac{-5}{5}\\\frac{dy}{dx}=-1[/tex]
Substitute m = -1 and the point (1, 1) into the equation of a line expressed as:
[tex]y-1=-1(x-1)\\y-1=-x+1\\y=-x+1+1\\y=-x+2[/tex]
Hence the equation of the tangent line to the curve at the given point is
y = -x = 2
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