Another implicit function problem

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SUMMARY

The discussion focuses on finding the tangent point between the ellipse defined by the equation \(9x^{2}+y^{2}=36\) and the hyperbola \(xy=a\) in the first quadrant. The derivatives for the ellipse and hyperbola are established as \(\frac{\partial y}{\partial x}=-9\frac{x}{y}\) and \(\frac{\partial y}{\partial x}=-\frac{y}{x}\), respectively. The solution involves substituting the hyperbola into the ellipse, leading to the equation \(9x^4-36x^2+36=0\). The discriminant condition yields \(a=6\) and the point of tangency is determined to be \((\sqrt{2}, 3\sqrt{2})\).

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Yankel
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Hello, I have one more (hopefully last) question regarding implicit functions:

The ellipse

\[9x^{2}+y^{2}=36\]

and the hyperbole

\[xy=a\]

tangent at a point in the first quarter.

I need to find the tangent point and a.

thanks !

I know that for the ellipse:

\[\frac{\partial y}{\partial x}=-9\frac{x}{y}\]

and for the hyperbole:

\[\frac{\partial y}{\partial x}=-\frac{y}{x}\]
 
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Solved it (Nod)

thanks anyway :o
 
A precalculus technique would be to solve the hyperbola for $y$ and then substitute into the ellipse to get in standard form:

$$9x^4-36x^2+a^2=0$$

For the two curves to be tangent, we require the discriminant to be zero:

$$36^2-4\cdot9\cdot a^2=0$$

$$a^2=36$$

Since a tangent point is to be in the first quadrant, we take the positive root:

$$a=6$$

Hence, we have:

$$9x^4-36x^2+36=0$$

$$x^4-4x^2+4=0$$

$$\left(x^2-2\right)^2=0$$

Taking the positive root:

$$x=\sqrt{2}\implies y=\frac{6}{\sqrt{2}}=3\sqrt{2}$$

And so the point of tangency is:

$$\left(\sqrt{2},3\sqrt{2}\right)$$

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