7.0 Pair of tangents

  Hyperbola

The equation of the pair of tangents which can be drawn from any point $({x_1},{y_1})$ to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is given by $$S{S_1} = {T^2}$$ where, $S \equiv \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} - 1;{\text{ }}{S_1} = \frac{{{x_1}^2}}{{{a^2}}} - \frac{{{y_1}^2}}{{{b^2}}} - 1;{\text{ }}T = \frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} - 1$

Question 7. Find the locus of point of intersection of perpendicular tangents to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$.

Solution: Let us assume $P(h,k)$ be the point of intersection of two perpendicular tangents. Equation of pair of tangents can be find out using $S{S_1} = {T^2}$ i.e., $$(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} - 1)(\frac{{{h^2}}}{{{a^2}}} - \frac{{{k^2}}}{{{b^2}}} - 1) = {(\frac{{hx}}{{{a^2}}} - \frac{{ky}}{{{b^2}}} - 1)^2}$$$$\frac{{{x^2}}}{{{a^2}}}( - \frac{{{k^2}}}{{{b^2}}} - 1) - \frac{{{y^2}}}{{{b^2}}}(\frac{{{h^2}}}{{{a^2}}} - 1) + ...... = 0...(1)$$ Since equation $(1)$ represents two perpendicular lines as given in question, $$\frac{1}{{{a^2}}}( - \frac{{{k^2}}}{{{b^2}}} - 1) - \frac{1}{{{b^2}}}(\frac{{{h^2}}}{{{a^2}}} - 1) = 0$$$$ - {k^2} - {b^2} - {h^2} + {a^2} = 0$$

The locus is $${x^2} + {y^2} = {a^2} - {b^2}$$

NOTE: The locus of point of intersection of tangents which are at right angle is known as the director circle. The equation of director circle is ${x^2} + {y^2} = {a^2} - {b^2}$.

In the above equation of director circle,

• If ${b^2} < {a^2}$, then the director circle is real.
• If ${b^2} = {a^2}$ (i.e., rectangular hyperbola), then the radius of the director circle is $0$ and it reduces to a point circle at the origin. Thus, centre is the only point from which two perpendicular tangents can be drawn.
• If ${b^2} > {a^2}$, then the radius of director circle is imaginary. Thus, there is no such circle and so no pair of tangents at right angle can be drawn.