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Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are $y=x+\sqrt{a^2-b^2},y=x-\sqrt{a^2-b^2},y=-x+\sqrt{a^2-b^2},y=-x-\sqrt{a^2-b^2}$.

I tried to solve the two equations and find the points of intersections of the hyperbolas but these hyperbolas do not intersect each other.

Let the point of tangency be $(x_1,y_1)$ on the $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $(x_2,y_2)$ on the $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{dy}{dx}=\frac{b^2x}{a^2y}$

For $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{dy}{dx}=\frac{a^2x}{b^2y}$

$\frac{b^2x_1}{a^2y_1}=\frac{a^2x_2}{b^2y_2}$

$b^4x_1y_2=a^4x_2y_1$

I do not know how to take it further.

diya
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2 Answers2

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Hint: Implicitly differentiate each, then set the derivatives equal to one another.

Leonidas Lanier
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Intersecting the dual conics is one way to go.

The dual conics are $$a^2X^2-b^2Y^2=1\quad (I)$$ and $$-b^2X^2+a^2Y^2=1\quad(II).$$ $b^2(I)+a^2(II)$ gives $(-b^4+a^4)Y^2=b^2+a^2$ or $(a^2-b^2)Y^2=1$. Substituting back to get the corresponding $X$s, we get the four intersection points to be $(X,Y)=(\frac1{\sqrt{a^2-b^2}},\pm\frac1{\sqrt{a^2-b^2}})$ and $(X,Y)=(-\frac1{\sqrt{a^2-b^2}},\pm\frac1{\sqrt{a^2-b^2}})$.

The four common tangents are $$Xx+Yy+1=0.$$

Jan-Magnus Økland
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