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प्रश्न
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
उत्तर
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.
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