A parabola, one of the most common forms of a quadratic function, is a curve where any point is at an equal distance from a fixed point (the focus), and a fixed straight line (the directrix). When a line touches this curve precisely at one point, it is referred to as a tangent to a parabola. This article aims to provide you with a comprehensive understanding of the equation of the tangent to a parabola and the point of contact of the tangent to a parabola, illustrated with examples.

Condition for Tangency

The line y = mx + c can be a tangent to the parabola y2 = 4ax if the condition c = a/m is met.

Maths Notes Free PDFs

Similarly, the line y = mx + c can be a tangent to the parabola x2 = 4ay if the condition c = -am2 is fulfilled.

The line x cos θ + y sin θ = p can be a tangent to the parabola y2 = 4ax, if the condition a sin2θ + p cos θ = 0 is satisfied.

1. Point Form:

The equation of a tangent to the parabola y2 = 4ax at a specific point P(x1, y1) is given by the formula yy1 = 2a(x + x1).

• Start with the parabola:
  y² = 4ax … (1)

• Since point P lies on the parabola:
  y₁² = 4ax₁ … (2)

• Differentiate (1) with respect to x:
  2y (dy/dx) = 4a ⇒ dy/dx = 2a/y

• At point (x₁, y₁), the slope m = 2a/y₁

• Use the point-slope form of a line:
  y – y₁ = m(x – x₁)
  ⇒ y – y₁ = (2a/y₁)(x – x₁)
  ⇒ Multiply both sides by y₁:
  yy₁ – y₁² = 2a(x – x₁)

• Now plug in equation (2):
  yy₁ – 4ax₁ = 2a(x – x₁)
  ⇒ Rearranging:
  yy₁ = 2a(x + x₁)

2. Slope Form

Let’s say the slope of the tangent is m.

a) If the parabola is y² = 4ax, the tangent is:

y = mx + a/m

• Here, the condition for a tangent is that c = a/m

• So, you write the equation of a line as y = mx + c

• Replace c with a/m ⇒ Final form: y = mx + a/m

• The point where the line touches the parabola is:
  (a/m², 2a/m)

b) If the parabola is x² = 4ay, the tangent is:

y = mx – am²

• The point of contact is:
  (2am, am²)

3. Parametric Form

The point on the parabola is written in terms of a parameter t as:
(at², 2at)

Then, the equation of the tangent at that point is:

ty = x + at²

• From the point form:
  yy₁ = 2a(x + x₁)

• Replace x₁ = at² and y₁ = 2at
  ⇒ y(2at) = 2a(x + at²)
  ⇒ Divide both sides by 2a:
  yt = x + at²

Key Points to Know When Finding a Tangent to a Parabola

1. Find the Slope (Gradient)
   To get the slope of the tangent, you take the derivative of the parabola. The derivative tells you how steep the curve is at any point.
    

2. Use the Point of Contact
   You need to know the exact point on the parabola where the tangent touches. This is often given as a coordinate or through a parameter like ttt.
    

3. Use the Slope-Point Formula
   Once you know the slope and the point of contact, you can use the formula for a straight line:
   y−y1=m(x−x1)

Solved Examples 

Example 1:

Find the equation of the tangent to the parabola y2 = 16x at the point where the parameter t = 3.

Options:
a. x – 3y + 18 = 0
b. x – 3x – 18 = 0
c. x + 3y + 18 = 0
d. none of these

Example 2:

Determine the condition that the line x cos θ + y sin θ = P touches the parabola y2 = 4ax.

Options:
a. P cos θ + a sin2 θ = 0
b. cos θ + a P sin2 θ = 0
c. P cos θ – a sin2 θ = 0
d. None of these

Further Reading

Conic Sections

JEE Previous Year Questions with Solutions on Parabola

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