Tangent to a Parabola

Let P(x₁, y₁) and Q(x₂, y₂) be two neighbouring points on the parabola y² – 4ax. Then the equation of the line joining P and Q is

   y – y₁ = (y₂-y₁) / (x₂-x₁ ) (x – x₁) …… (1)

Since, points P and Q lies on the parabola, we have

   y₁² = 4ax₁ …… (2)

y₂² = 4ax₂ …… (3)

   (equation 3 and 2) give

   y₂² – y₁² = 4a(x₂ – x₁)

   ⇒ (y₂-y₁)/(x₂-x₁ )=4a/(y₁+y₂ )

   Equation of chord PQ (i.e. equation (1) becomes):

y – y₁ = 4a/(y₁+y₂ ) (x – x₁) …… (4)

Our aim is to in the equation of tangent at point P. For that, let point Q approach point P i.e. x₂ → x₁ and y₂ → y₁.

   y – y₁ = 4a/(2y₁ ) (x – x₁)

   ⇒ y₁y = 2a (x₁ + x) (using equation (2))

This is the required equation of the tangent to the parabola y² – 4ax at P(x₁, y₁).

Note:

The angle between the tangents drawn to the two parabolas at the point of their intersection is defined as the angle of intersection of two parabolas.

Tangent at the point (x₁, y₁)

   Let the equation of the parabola be y² = 4ax.

   Hence, value of dy/dx at P(x₁, y₁) is 2a/y₁ and the equation of the tangent at P is

y – y₁ = 2a/y₁ (x – x₁) i.e. yy₁ = 2a(x – x₁) + y₁².

⇒ yy₁ = 2a(x + x₁).

   Alternatively, we write the equation of the chord joining the points P(x₁, y₁) and Q(x₂, y₂) on the parabola y² = 4ax. Equation of the chord is (x-x₁)/(x₂-x₁ )=(y-y₁)/(y₂-y₁ )

or (x-x₁)/(x₂-x₁ ) = (y-y₁ )(y₂+y₁ )/(y₂²-y₁² )= (y-y₁ )(y₂+y₁ )/((x₂-x₁ ) )

or 4a(x – x) = (y – y₁) (y₂ + y₁).

When the two points P and Q tend to coincide, y₂ → y₁ and the line PQ becomestangent to the parabola. Its equation is 4a (x – x₁) = (y – y₁) (2y₁) = 2yy₁ – 2y₁²= 2yy₁ – 8ax₁ or yy₁ = 2a(x + x₁).

Tangent in terms of m

Suppose that the equation of a tangent to the parabola y² = 4ax … (i)

is y = mx + c. … (ii)

   The abscissae of the points of intersection of (i) and (ii) are given by the equation (mx + c)² = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points

   ⇒ (mx – 2a)² = m²c² … (iii)

⇒ c = a/m. … (iv)

Hence, y = mx + a/m is a tangent to the parabola y² = 4ax, whatever be the value of m.

   Equation (mx + c)² = 4ax now becomes (mx – a/m)² = 0.

   ⇒ x = a/m² and y² = 4ax ⇒ y = 2a/m.

   Thus the point of contact of the tangent y = mx + a/m is (a/m² ,2a/m).

Tangent to a Parabola

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