P and Q are two points on the ellipse `x^2/"a"^2 + y^2/"b"^2` = 1 with eccentric angles θ_{1} and θ_{2}. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ_{1} + θ_{2} = `π/2`

#### Solution

Given equation of the ellipse is `x^2/"a"^2 + y^2/"b"^2` = 1

θ_{1} and θ_{2} are the eccentric angles of tangent.

∴ Equation of tangent at point P is

`x/"a"costheta_1 + y/"b"sintheta_1` = 1 ...(i)

∴ Equation of tangent at point Q is

`x/"a"costheta_2 + y/"b"sintheta_2` = 1

θ_{1} + θ_{2} = `π/2` ...[Given]

∴ θ_{2} = `pi/2 - theta_1`

`x/"a"cos(pi/2 - theta_1) + y/"b"sin(pi/2 - theta_1)` = 1

∴ `x/"a"sintheta_1 + y/"b"costheta_1` = 1 ...(ii)

From (i) and (ii), we get

`x/"a"costheta_1 + y/"b"sintheta_1 = x/"a"sintheta_1 + y/"b"costheta_1`

Let M(x_{1}, y_{1}) be the point of intersection of the tangents.

∴ `x_1/"a" costheta_1 + y_1/"b" sintheta_1 = x_1/"a" sintheta_1 + y_1/"b"costheta_1`

∴ `x_1/"a"(costheta_1 - sintheta_1) = y_1/"b"(costheta_1 - sintheta_1)` ...(iii)

If cos θ_{1} – sin θ_{1} = 0,

cosθ_{1 }= sin θ_{1}

∴ tan θ_{1} = 1

∴ θ_{1} = `pi/4`

Since θ_{1} + θ_{2} = `pi/2`, θ_{2} = `pi/2 - pi/4 = pi/4`

i.e., points P and Q coincide, which is not possible, as P and Q are two different points.

∴ cos θ_{1} – sin θ_{1} ≠ 0

Dividing equation (iii) by (cos θ_{1} – sin θ_{1}), we get

`x_1/"a" = y_1/"b"`

∴ bx_{1} – ay_{1} = 0

∴ bx – ay = 0, which is the required equation of locus of point M.