Analytic Geometry – Problem 33: find and the equation of

Consider the hyperbola $C: x^2-\dfrac{y^2}{b^2}=1\,(b>0)$. Its asymptotes are two straight lines through the origin. The circle $x^2+y^2=1$ intersects these two asymptotes at four points, forming a rectangle.

If the area of this rectangle is $2b$, find $b$ and the equation of $C$.

(1) The asymptotes of $x^2-\dfrac{y^2}{b^2}=1$ are $$y=\pm bx.$$ (2) Intersect $y=bx$ with $x^2+y^2=1$: $$x^2+(bx)^2=1\Rightarrow x^2(1+b^2)=1\Rightarrow x_0=\frac{1}{\sqrt{1+b^2}},\quad y_0=\frac{b}{\sqrt{1+b^2}}.$$ By symmetry, the four intersection points are $(\pm x_0,\pm y_0)$, which form a rectangle of side lengths $2x_0$ and $2y_0$. Hence area $$S=4x_0y_0=4\cdot\frac{1}{\sqrt{1+b^2}}\cdot\frac{b}{\sqrt{1+b^2}}=\frac{4b}{1+b^2}.$$ (3) Given $S=2b$, we have $$\frac{4b}{1+b^2}=2b.$$ Since $b>0$, divide by $2b$: $\frac{2}{1+b^2}=1\Rightarrow 1+b^2=2\Rightarrow b^2=1\Rightarrow b=1$.

Thus $C$ is $$x^2-y^2=1.$$

We get $b=1$, so the hyperbola is $x^2-y^2=1$.