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# P and Q are two points on the ellipse x2a2+y2b2 = 1 with eccentric angles θ1 and θ2. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ1 + θ2 = ππ2 - Mathematics and Statistics

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P and Q are two points on the ellipse x^2/"a"^2 + y^2/"b"^2 = 1 with eccentric angles θ1 and θ2. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ1 + θ2 = π/2

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#### Solution

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

θ1 and θ2 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

θ1 + θ2 = π/2    ...[Given]

∴ θ2 = 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(x1, y1) 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 θ1 – sin θ1 = 0,

cosθ= sin θ1

∴ tan θ1 = 1

∴ θ1 = pi/4

Since θ1 + θ2 = pi/2,  θ2 = 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 θ1 – sin θ1 ≠ 0

Dividing equation (iii) by (cos θ1 – sin θ1), we get

x_1/"a" = y_1/"b"

∴ bx1 – ay1 = 0

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

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