Let PQ$PQ$ be a chord of the hyperbola fracx^24-fracy^2b^2=1$\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1$, perpendicular to the x-axis such that OPQ$OPQ$ is an equilateral triangle, O$O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is sqrt3$\sqrt{3}$, then the area of the triangle OPQ$OPQ$ is:

### Related Formula Eccentricity of a hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is e = sqrt1 + fracb^2a^2$e = \sqrt{1 + \frac{b^2}{a^2}}$ Parametric coordinates on a hyperbola are (asectheta, btantheta)$(a\sec\theta, b\tan\theta)$. ### Core Logic

Given e = sqrt3$e = \sqrt{3}$ and a^2 = 4$a^2 = 4$ (a=2$a=2$): e^2 = 1 + fracb^24 implies 3 = 1 + fracb^24 implies b^2 = 8 implies b = 2sqrt2$$e^2 = 1 + \frac{b^2}{4} \implies 3 = 1 + \frac{b^2}{4} \implies b^2 = 8 \implies b = 2\sqrt{2}$$ The equation is fracx^24 - fracy^28 = 1$\frac{x^2}{4} - \frac{y^2}{8} = 1$. Let the point P$P$ on the hyperbola be (2sectheta, 2sqrt2tantheta)$(2\sec\theta, 2\sqrt{2}\tan\theta)$. Since chord PQ$PQ$ is perpendicular to the x-axis, the triangle OPQ$OPQ$ has the x-axis as its altitude OM$OM$. In equilateral triangle OPQ$\triangle OPQ$, half the vertex angle is 30^circ$30^\circ$: tan 30^circ = fracPMOM$$\tan 30^\circ = \frac{PM}{OM}$$ ### Step 1: Calculating Coordinates frac1sqrt3 = frac2sqrt2tantheta2sectheta = sqrt2sintheta$$\frac{1}{\sqrt{3}} = \frac{2\sqrt{2}\tan\theta}{2\sec\theta} = \sqrt{2}\sin\theta$$ sintheta = frac1sqrt6$$\sin\theta = \frac{1}{\sqrt{6}}$$ ### Step 2: Area Calculation The area of triangle OPQ = 2 times left(frac12 times OM times PMright) = OM times PM$\triangle OPQ = 2 \times \left(\frac{1}{2} \times OM \times PM\right) = OM \times PM$ textArea = (2sectheta) times (2sqrt2tantheta) = 4sqrt2 fracsinthetacos^2theta$$\text{Area} = (2\sec\theta) \times (2\sqrt{2}\tan\theta) = 4\sqrt{2} \frac{\sin\theta}{\cos^2\theta}$$ Since sintheta = frac1sqrt6$\sin\theta = \frac{1}{\sqrt{6}}$, we have cos^2theta = 1 - frac16 = frac56$\cos^2\theta = 1 - \frac{1}{6} = \frac{5}{6}$. textArea = 4sqrt2 fracfrac1sqrt6frac56 = 4sqrt2 left(frac1sqrt6right)left(frac65right) = 4sqrt2 left(fracsqrt65right) = frac4sqrt125 = frac8sqrt35$$\text{Area} = 4\sqrt{2} \frac{\frac{1}{\sqrt{6}}}{\frac{5}{6}} = 4\sqrt{2} \left(\frac{1}{\sqrt{6}}\right)\left(\frac{6}{5}\right) = 4\sqrt{2} \left(\frac{\sqrt{6}}{5}\right) = \frac{4\sqrt{12}}{5} = \frac{8\sqrt{3}}{5}$$ ### Pattern Recognition A perpendicular chord symmetric about the principal axis automatically splits into two right triangles. Utilizing the parametric coordinates directly defines the ratio of sides matching tan 30^circ$\tan 30^\circ$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Hyperbola