Hyperbola
Hyperbola
nta_abhyas_2025
Grade 11
Question:
Given hyperbola is $3x^2 - 2y^2 = 6$ or $\frac{x^2}{2} - \frac{y^2}{3} = 1$. Slope form of tangent is $y = mx \pm \sqrt{a^2m^2 - b^2}$ or $(mx - y)^2 = a^2m^2 - b^2$. Tangent from the point $(\alpha, \beta)$ is given by, $(\beta - m\alpha)^2 = 2m^2 - 3$, i.e., $m^2(\alpha^2 - 2) - 2\alpha m\beta + \beta^2 + 3 = 0$, so $m_1m_2 = \frac{\beta^2 + 3}{\alpha^2 - 2} = \tan\theta \tan\phi$.
Step-by-Step Solution
Key Concept: The product of slopes of tangents from an external point relates to the parameters of the hyperbola and the coordinates of the point.
For the hyperbola $3x^2 - 2y^2 = 6$ which simplifies to $\frac{x^2}{2} - \frac{y^2}{3} = 1$, the slope form of tangent is $y = mx \pm \sqrt{2m^2 - 3}$. The tangent from point $(\alpha, \beta)$ satisfies $(\beta - m\alpha)^2 = 2m^2 - 3$, giving the quadratic $m^2(\alpha^2 - 2) - 2\alpha m\beta + \beta^2 + 3 = 0$. The product of slopes is $m_1m_2 = \frac{\beta^2 + 3}{\alpha^2 - 2}$, which equals $\tan\theta\tan\phi$ for the angle between the tangents.
Correct Answer: 7