Answer: At the stationary points, \(\frac{dy}{dx}=0\). Differentiate \(x^{2}+4xy-y^{2}+20=0\) implicitly:

\(2x+4\left(x\frac{dy}{dx}+y\right)-2y\frac{dy}{dx}=0\).

Setting \(\frac{dy}{dx}=0\) gives \(2x+4y=0\), so \(x=-2y\).

Substitute this into the equation of the curve:

\(( -2y)^2+4(-2y)(y)-y^2+20=0\)

\(4y^2-8y^2-y^2+20=0\)

\(5y^2=20\), so \(y=\pm 2\).

Hence the stationary points are \((4,-2)\) and \((-4,2)\).

To determine their nature, differentiate the first derivative equation again:

\(2x+4\left(x\frac{dy}{dx}+y\right)-2y\frac{dy}{dx}=0\)

gives

\(2+4\left(x\frac{d^2y}{dx^2}+2\frac{dy}{dx}\right)-2\left(y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right)=0\).

At a stationary point, \(\frac{dy}{dx}=0\), so this simplifies to

\(2+(4x-2y)\frac{d^2y}{dx^2}=0\).

For \((4,-2)\): \(2+(16+4)\frac{d^2y}{dx^2}=0\), so \(\frac{d^2y}{dx^2}=-\frac{1}{10}\). This is a maximum.

For \((-4,2)\): \(2+(-16-4)\frac{d^2y}{dx^2}=0\), so \(\frac{d^2y}{dx^2}=\frac{1}{10}\). This is a minimum.

Let \(y' = \frac{dy}{dx}\).

Differentiate \(x^2+4xy-y^2+20=0\) implicitly:

\(2x+4(xy' + y)-2yy' = 0\).

At a stationary point, \(y'=0\), so

\(2x+4y=0\)

and hence \(x=-2y\).

Now substitute \(x=-2y\) into the curve equation:

\(( -2y)^2 + 4(-2y)(y) - y^2 + 20 = 0\)

\(4y^2 - 8y^2 - y^2 + 20 = 0\)

\(-5y^2 + 20 = 0\)

\(5y^2 = 20\)

\(y^2 = 4\)

so \(y=\pm 2\).

Using \(x=-2y\):

• if \(y=-2\), then \(x=4\), giving \((4,-2)\);
• if \(y=2\), then \(x=-4\), giving \((-4,2)\).

So the stationary points are \((4,-2)\) and \((-4,2)\).

To find their nature, differentiate \(2x+4(xy'+y)-2yy'=0\) again:

\(2+4(xy''+y'+y')-2(yy''+(y')^2)=0\),

so

\(2+4xy''+8y' -2yy'' -2(y')^2=0\).

At a stationary point, \(y'=0\), hence

\(2+(4x-2y)y''=0\).

At \((4,-2)\):

\(2+(16+4)y''=0\Rightarrow 20y''=-2\Rightarrow y''=-\frac{1}{10}\).

Since \(y''\lt 0\), this point is a maximum.

At \((-4,2)\):

\(2+(-16-4)y''=0\Rightarrow -20y''=-2\Rightarrow y''=\frac{1}{10}\).

Since \(y''\gt 0\), this point is a minimum.