Equation of a Chord joining two points

In this article, we discuss the equation of a chord joining two points on an ellipse, hyperbola and circle. As far as any entrance exam is concerned, this is an important topic. Students can expect one question from this topic.

The line joining any two points on the circle is called the chord of the circle. The equation of the chord joining two points on an ellipse, hyperbola and circle is given below.

Formula

• Equation of the chord joining two points having eccentric angles α and β on the ellipse x²/a² + y²/b² = 1 is (x/a) cos (α+β)/2 + (y/b) sin (α+β)/2 = cos (α-β)/2.
• For a hyperbola, the equation of the chord joining two points P(a sec α, b tan α) and Q(a sec β, b tan β) is (x/a) cos (α-β)/2 – (y/b) sin (α+β)/2 = cos (α+β)/2.
• The equation of chord joining P(α) = (a cos α, a sin α) and Q(β) = (a cos β, a sin β) on the circle x² + y² = a² is (x/a) cos (α+β)/2 + (y/a) sin (α+β)/2 = cos (α-β)/2.

Also Read

Important Properties of Hyperbola

Conic Sections

Solved Examples

Example 1:

The equation of the chord joining two points (x₁,y₁) and (x₂,y₂) on the rectangular hyperbola xy = c² is

1) x/(x₁ + x₂) + y/(y₁ + y₂) = 1

2) x/(x₁ – x₂) + y/(y₁ – y₂) = 1

3) x/(y₁ + y₂) + y/(x₁ + x₂) = 1

4) x/(y₁ – y₂) + y/(x₁ – x₂) = 1

Solution:

Given hyperbola xy = c²

The points (x₁,y₁) and (x₂,y₂) lie on it.

So x₁y₁ = c²

y₁ = c²/x₁

(x₁, y₁) = (x₁, c²/x₁)

Similarly x₂y₂ = c²

y₂ = c²/x₂

(x₂, y₂) = (x₂, c²/x₂)

Slope of chord, m = c²(x₁-x₂)/x₁x₂(x₂-x₁)

= -c²/x₁x₂

Equation of chord is given by y – y₁ = m(x-x₁)

=> (y-y₁)/(x-x₁) = -c²/x₁x₂

=> x₁x₂y – x₁x₂y₁ = -c²x + c²x₁

=> x₁x₂y – c²x₂ = -c²x + c²x₁

=> x₁x₂y + c²x = c²(x₂ + x₁)

=> [x₁x₂y/c²(x₂ + x₁)] + c²x/c²(x₂ + x₁) = 1…(ii)

Also y₁ + y₂ = (c²/x₁) + (c²/x₂)

= c²(x₁ + x₂)/x₁x₂

So from (ii)

y/(y₁+y₂) + x/(x₁ + x₂) = 1

Hence, option 1 is the answer.

Example 2:

The chords joining the points θ₁ and θ₂ on the ellipse x²/a² + y²/b² = 1 intersect the major-axis in (h, 0). Prove that tan(θ₁/2) tan(θ₂/2) = (h-a)/(h+a)

Solution:

Equation of the chord joining θ₁ and θ₂ is

(x/a) cos (θ₁+ θ₂)/2 + (y/b) sin ( θ₁+ θ₂)/2 = cos ( θ₁– θ₂)/2

Chords pass through (h, 0).

=> (h/a) cos (θ₁+ θ₂)/2 + 0 = cos (θ₁– θ₂)/2

=> h/a = [cos (θ₁– θ₂)/2]/[cos (θ₁+ θ₂)/2]

=> (h-a)/(h+a) =[cos (θ₁– θ₂)/2 – cos (θ₁+ θ₂)/2]/[cos (θ₁– θ₂)/2 + cos (θ₁+ θ₂)/2]

= tan(θ₁/2) tan (θ₂/2)

Hence proved.

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