5.1 Point form

  Hyperbola

The equation of tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ at a point $P({x_1},{y_1})$ can be find out using $T=0$ i.e., $$\frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} = 1$$

Question 3. A line $2x + \sqrt 6 y = 2$ touches the hyperbola ${x^2} - 2{y^2} = 4$. Find the point of contact.

Solution: Let us assume the point of contact be $({x_1},{y_1})$. Equation of tangent using point form is $$x{x_1} - 2y{y_1} = 4$$$$\frac{{x{x_1}}}{2} - y{y_1} = 2$$ On comparing it with $$2x + \sqrt 6 y = 2$$ We get, $$\frac{{{x_1}}}{2} = 2{\text{ and }}{y_1} = - \sqrt 6 $$Therefore, the point of contact is $$({x_1},{y_1}) \equiv (4, - \sqrt 6 )$$