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प्रश्न
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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उत्तर
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,
cos θ1 = 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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