Question

Show that the locus of the point of intersection of tangents at two points on an ellipse, whose eccentric angles differ by a constant, is an ellipse

Answer 1

Let the equation of the ellipse be `x^2/"a"^2 + y^2/"b"^2` = 1.

Let P(θ₁) and Q(θ₂) be any two points on the ellipse such that

θ₂ – θ₁ = k  ...(k is a constant) ...(1)

∴ P = `("a"  costheta_1, "b"  sintheta_1)`

Q = `("a"  costheta_2, "b"  sintheta_2)`

The equations of tangent to the ellipse at P is

`(x("a"  costheta_1))/"a"^2 + (y("b"  sintheta_1))/"b"^2` = 1

∴ `(x  cos  theta_1)/"a" + (y  sintheta_1)/"b"` = 1

∴ bx cos θ₁ + ay sin θ₁ = ab ...(2)

Similarly, the equation of tangent at Q is

bx cos θ₂ + ay sin θ₂ = ab  ...(3)

We solve equations (2) and (3).

Multiplying equation (2) by sin θ₂ and equation (3) by sin θ₁, we get,

bx sin θ₂ cos θ₁ + ay sin θ₁ sin θ₂ = ab sin θ₂

and bx sin θ₁ cos θ₂ + ay sin θ₁ sin θ₂ = ab sin θ₁

On subtracting, we get,

bx (sin θ₁ cos θ₂ – cos θ₁ sin θ₂) = ab(sin θ₁ – sin θ₂)

∴ bx sin (θ₁ – θ₂) = ab (sin θ₁ – sin θ₂)

∴ `"b"x xx 2sin((theta_1 - theta_2)/2) cos((theta_1 - theta_2)/2) = "ab" xx 2cos((theta_1 + theta_2)/2)sin((theta_1 - theta_2)/2)`

∴ `bxcos((theta_1 - theta_2)/2) = "ab"cos((theta_1 + theta_2)/2)`

∴ `"b"xcos(-"k"/2) = "ab"cos((theta_1 + theta_2)/2)`  ...[By (1)]

∴ x = `("a"cos((theta_1 + theta_2)/2))/cos("k"/2)`  ...[∵ cos(– θ) = cos θ]

Again, multiplying equation (2) by cos θ₂ and equation (3) by cos θ₁, we get,

bx cos θ₁ cos θ₂ + ay sin θ₁ cos θ₂ = ab cos θ₂

and bx cos θ₁ cos θ₂ + ay cos θ₁ sin θ₂ = ab cos θ₁

On subtracting, we get,

ay (sin θ₁ cos θ₂ – cos θ₁ sin θ₂) = ab (cos θ₂ – cos θ₁)

∴ ay sin (θ₁ – θ₂) = ab (cos θ₂ – cos θ₁)

∴ `"a"y xx 2sin((theta_1 - theta_2)/2) cos((theta_1 - theta_2)/2) = "ab" xx 2sin((theta_1 + theta_2)/2)sin((theta_1 - theta_2)/2)`

∴ `ay cos((theta_1 - theta_2)/2) = "ab"sin((theta_1 + theta_2)/2)`

∴ `"a"y cos(-"k"/2) = "ab"sin((theta_1 + theta_2)/2)`  ...[By (1)]

∴ y = `("b"sin((theta_1 + theta_2)/2))/cos("k"/2)`  ...[∵ cos(– θ) = cos θ]

If R = ( x₁, y₁) is any point on the required locus, then R is the point of intersection of tangents at P and Q.

∴ x₁ = `("a"cos((theta_1 + theta_2)/2))/cos("k"/2)`, y₁ = `("b"sin((theta_1 + theta_2)/2))/cos("k"/2)`

∴ `cos((theta_1 + theta_2)/2) = (x_1 cos("k"/2))/"a"` and 

`sin((theta_1 + theta_2)/2) = (y_1cos("k"/2))/"b"`

But `cos^2  ((theta_1 + theta_2))/2 + sin^2  ((theta_1 + theta_2))/2` = 1

∴ `x_1^2 cos^2("k"/2)^2/"a"^2 + (y_1^2 cos^2("k"/2)^2)/"b"^2` = 1

Replacing x₁ by x and y₁ by y, the equation of the required locus is

∴ `x^2/("a"^2 sec^2("k"/2)) + y^2/("b"^2 sec^2("k"/2))` = 1

which is an ellipse.