Tangents are drawn from the point (α, β) to the hyperbola 3 x ^ 2 -...

Given hyperbola is 3{x}^{2} - 2{y}^{2} = 6 or $\dfrac{{x}^{2}}{2} - \dfrac{{y}^{2}}{3} = 1$
Slope form of tangent is $y = mx \pm \sqrt{{a}^{2}{m}^{2} - {b}^{2}}$or ${\left(y - mx\right)}^{2} = {a}^{2}{m}^{2} - {b}^{2}$

Tangent from the point \left(\alpha, \beta\right)$is given by,</p><p>${\left(\beta - m\alpha\right)}^{2} = 2{m}^{2} - 3$<br />$⇒ {m}^{2}\left({\alpha}^{2} - 2\right) - 2\alpha\betam + {\beta}^{2} + 3 = 0$<br />${m}_{1}{m}_{2} = \dfrac{{\beta}^{2} + 3}{{\alpha}^{2} - 2} = 2 = tan \theta tan ϕ$<br />$∴ {\beta}^{2} + 3 = 2{\alpha}^{2} - 4 ⇒ 7 = 2{\alpha}^{2} - {\beta}^{2}$