Equation of Pair of Tangents

When studying conic sections such as hyperbolas, an important concept is the equation of a pair of tangents drawn from a point to the conic. This concept is crucial for understanding the geometric properties of hyperbolas and solving related problems in exams.

Understanding the Pair of Tangents

A pair of tangents can be drawn from an external point to a hyperbola. These tangents are significant because they reveal information about the conic's structure and the relationship between the point and the hyperbola.

Standard Equation of a Hyperbola

The standard equation of a hyperbola centered at the origin, with its transverse axis along the x-axis and conjugate axis along the y-axis, is given by:

$$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $$

where (a) is the semi-major axis and (b) is the semi-minor axis.

Equation of a Tangent to a Hyperbola

For a point ( (x_1, y_1) ) lying on the hyperbola, the equation of the tangent to the hyperbola at this point is:

$$ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 $$

Equation of Pair of Tangents from an External Point

Let ( (x_0, y_0) ) be an external point from which two tangents are drawn to the hyperbola. The equation of the pair of tangents from this point can be derived using the concept of homogenization.

The combined equation of the pair of tangents from the point ( (x_0, y_0) ) to the hyperbola ( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 ) is given by:

$$ T^2 = SS_1 $$

where ( T ) is the equation of the tangent, ( S ) is the equation of the hyperbola, and ( S_1 ) is the equation of the hyperbola with ( (x, y) ) replaced by ( (x_0, y_0) ).

Expanding this, we get:

$$ \left( \frac{x x_0}{a^2} - \frac{y y_0}{b^2} \right)^2 = \left( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} - 1 \right) \left( \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1 \right) $$

This equation represents the pair of tangents drawn from the external point ( (x_0, y_0) ) to the hyperbola.

Important Points and Differences

Point/Difference	Tangent to Hyperbola	Pair of Tangents from External Point
Equation Form	Linear	Quadratic
Number of Tangents	One per point on the hyperbola	Two from an external point
Point of Contact	Lies on the hyperbola	No direct contact, tangents meet at the external point
Condition for Existence	Point must lie on the hyperbola	External point must not lie inside the conic

Examples

Example 1: Equation of a Tangent

Given the hyperbola ( \frac{x^2}{16} - \frac{y^2}{9} = 1 ) and a point ( (4, 3) ) on it, find the equation of the tangent at that point.

Using the formula for the tangent at ( (x_1, y_1) ), we get:

$$ \frac{x \cdot 4}{16} - \frac{y \cdot 3}{9} = 1 $$

Simplifying, we obtain:

$$ \frac{x}{4} - \frac{y}{3} = 1 $$

Example 2: Equation of Pair of Tangents

Find the equation of the pair of tangents drawn from the point ( (8, 6) ) to the hyperbola ( \frac{x^2}{16} - \frac{y^2}{9} = 1 ).

Using the formula for the pair of tangents, we have:

$$ \left( \frac{x \cdot 8}{16} - \frac{y \cdot 6}{9} \right)^2 = \left( \frac{8^2}{16} - \frac{6^2}{9} - 1 \right) \left( \frac{x^2}{16} - \frac{y^2}{9} - 1 \right) $$

Expanding and simplifying this equation will give us the quadratic equation representing the pair of tangents from the point ( (8, 6) ) to the given hyperbola.

Understanding the equation of a pair of tangents is essential for solving problems related to tangency and conics in exams. It requires a good grasp of algebraic manipulation and geometric interpretation.