**Intersection of a Line and Hyperbola**

Equation of line is y = mx + c . . . (1)

and equation of hyperbola is x²/a² – y²/b² = 1 . . . (2)

from equation (1) and (2)

b²x² – a² (mx + c) ² = a² b²

(b² – a²m²) x² – 2a²mcx – a² (b² + c²) = 0 . . . (3)

Hence y = mx ± √ (a²m² – b²) is a tangent to the standard hyperbola.

In this equation of tangent.

a²m² – b² ≥ 0

m² ≥ b²/a²

m ϵ (-∞, -b/a] È [b/a, ∞)

Hence, for given m, there can be two parallel tangents to the hyperbola.

If the tangents pass through (h, k), then

(k – m h) ² = a²m² – b²

(h² – a²) m² – 2kmh + k² + b² = 0

Hence, passing through a given point (h, k), there can be maximum of two tangents. Now

m₁ + m₂ = 2kh/ (h² – a²) . . . (4)

m₁ m₂ = (k² + b²)/ (h² – a²) . . . (5)

β is the angle enclosed by the tangent as shown in figure.

Now

tan²β = ((m₁ + m₂)² – 4m₁m₂) / (1 + m₁ m₂) . . . (6)

Equation (4) and (5) substitute in equation (6).

If β = 90⁰, the m₁ m₂ = -1

Hence, from (5), we get

k² + b² = a² – h²

x² + y² = a² – b²

which is the equation of the director circle of the hyperbola.

- If a > b the director circle is real is a point with finite radius.
- If a = b, the director circle is a point circle which is the origin
- If a < b, there no real circle or no such point on the plane.

**Note: **

For hyperbola (x – h)²/a² – (y – k) ²/b² = 1, the equation of tangent at point P(x₁, y₁) is

y – y₁ = m(x – x₁) ± √(a²m² – b²).

**Example:** For all real values of m, the straight-line y = mx + 6 is a tangent to
the hyperbola

x²/ 100 – y²/49 = 1.

**Solution:** Given that

straight-line y = mx + 6 and

hyperbola x²/ 100 – y²/49 = 1.

c² = a²m² – b²

c = 6, a² = 100 and b² = 49

Therefore, 36 = 100m² – 49

100m² = 85

m = √ (17/20)