Mathematics - Hyperbola Question with Solution | TestHub

If OPQ is an equilateral triangle, then OP makes an angle of $30^{\circ }$ with the x-axis.

The point $\left(\frac{\sqrt{3}r}{2},\frac{r}{2}\right)$ lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

Substituting these coordinates into the hyperbola equation, we get:

$$\frac{{\left(\frac{\sqrt{3}r}{2}\right)}^2}{a^2}-\frac{{\left(\frac{r}{2}\right)}^2}{b^2}=1$$

$$\frac{3r^2}{4a^2}-\frac{r^2}{4b^2}=1$$

$$r^2\left(\frac{3}{4a^2}-\frac{1}{4b^2}\right)=1$$

$$r^2\left(\frac{3b^2-a^2}{4a^2{b}^2}\right)=1$$

$$r^2=\frac{4a^2{b}^2}{3b^2-a^2}$$

For $r^2$ to be positive, we must have $3b^2-a^2>0$.

$$3b^2>a^2$$

$$\frac{b^2}{a^2}>\frac{1}{3}$$

Since $e^2=1+\frac{b^2}{a^2}$ for a hyperbola, we can substitute the inequality:

$$e^2-1>\frac{1}{3}$$

$$e^2>1+\frac{1}{3}$$

$$e^2>\frac{4}{3}$$

Taking the square root of both sides, we find the eccentricity $e$:

$$e>\sqrt{\frac{4}{3}}$$

$$e>\frac{2}{\sqrt{3}}$$