Question

The eccentric angles of two points P and Q the ellipse 4x² + y² = 4 differ by `(2pi)/3`. Show that the locus of the point of intersection of the tangents at P and Q is the ellipse 4x² + y² = 16

Answer 1

Given equation of the ellipse is 4x² + y² = 4

∴ `x^2/1 + y^2/4` = 1

Let P(θ₁) and Q(θ₂) be any two points on the given ellipse such that θ₁ – θ₂ = `(2pi)/3`

Equation of tangent at point P(θ₁) is

`(xcostheta_1)/1 + (ysintheta_1)/2` = 1   ...(i)

Equation of tangent at point Q(θ₂) is 

`(xcostheta_2)/1 + (ysintheta_2)/2` = 1   ...(ii)

Multiplying equation (i) by cos θ₂ and equation (ii) by cos θ₁ and subtracting, we get

`y/2(sintheta_1 costheta_2  -  sintheta_2 costheta_1)` = cos θ₂ – cos θ₁ 

∴ `y/2[sin(theta_1 - theta_2)]` = cos θ₂ – cos θ₁ 

∴ `y/2[sin((2pi)/3)]` = cos θ₂ – cos θ₁ 

∴ `y/2 sin(pi - pi/3)` = cos θ₂ – cos θ₁ 

∴ `y/2sin(pi/3)` = cos θ₂ – cos θ₁ 

∴ `y/2(sqrt(3)/2)` = cos θ₂ – cos θ₁ 

∴ `(sqrt(3)y)/4` = cos θ₂ – cos θ₁    ...(iii)

Multiplying equation (i) by sin θ₂ and equation (ii) by sin θ₁ and subtracting, we get

x(sin θ₂ cos θ₁ – cos θ₂ sin θ₁) = sin θ₂ – sin θ₁

∴ – x sin (θ₁ – θ₂) = sin θ₂ – sin θ₁

∴ `-xsin((2pi)/3)` = sin θ₂ – sin θ₁ 

∴  `-xsin(pi - pi/3)` = sin θ₂ – sin θ₁ 

∴ `-x sin  pi/3` = sin θ₂ – sin θ₁ 

∴ `- sqrt(3)/2x` = sin θ₂ – sin θ₁    ...(iv)

Squaring (iii) and (iv) and adding, we get

`(3x^2)/4 + (3y^2)/16` = sin² θ₂ – 2 sin θ₂ sin θ₁ + sin² θ₁ + cos² θ₂ – 2 cos θ₂ cos θ₁ + cos² θ₁

= (cos² θ₂ + sin² θ₂) + (cos² θ₁ + sin² θ₁) – 2 cos θ₂ cos θ₁ – 2 sin θ₂ sin θ₁

= 1 + 1 – 2 (cos θ₂ cos θ₁ + sin θ₂ sin θ₁)

= 2 – 2 [cos (θ₁ – θ₂)]

= `2 - 2cos((2pi)/3)`

= `2 - 2((-1)/2)`

= 2 + 1

∴ `(3x^2)/4 + (3y^2)/16` = 3

∴ `x^2/4 + y^2/16` = 1

∴ 4x² + y² = 16, which is the required equation of locus.