At any point P on the parabola y 2 – 2y – 4x + 5 =0, a tangent is drawn which meets the directrix at Q . Find the locus of R which divides QP externally in the ratio ½ : 1 .

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Latika Leekha · Answer

The given parabola isy 2 - 2y &ndash; 4x + 5 = 0. This can be rewritten as (y-1)2 = 4(x-1) Its parametric coordinates are x-1 = t 2 and y -1 = 2t and hence we have P(1 +t 2 , 1 + 2t) Hence, the equation of tangent at P is t(y-1) = x &ndash; 1 + t 2 , which meets the directrix x = 0 at Q. Hence, y = 1 + t &ndash; 1/t or Q(0, 1 + t &ndash; 1/t) Let R(h, k) be the point which divides QP externally in the ratio 1/2:1. Q is the mid-point of RP So, we have 0 = (h + t 2 + 1)/2 which gives t 2 = &ndash; (h + 1) &hellip;. (1) and 1 + t &ndash; 1/t = (k + 2t + 1)/2 which gives t = 2/(1- k) &hellip;.. (2) Hence, from equations (1) and (2) we get, 4/(1-k) 2 + (h + 1) = 0 or (k-1) 2 (h +1) + 4 = 0 Hence, the locus of a point is (x + 1)(y - 1) 2 = 0.

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At any point P on the parabola y 2 – 2y – 4x + 5 =0, a tangent is drawn which meets the directrix at Q . Find the locus of R which divides QP externally in the ratio ½ : 1 .

At any point P on the parabola y² – 2y – 4x + 5 =0, a tangent is drawn which meets the directrix at Q . Find the locus of R which divides QP externally in the ratio
½ : 1 .

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At any point P on the parabola y 2 – 2y – 4x + 5 =0, a tangent is drawn which meets the directrix at Q . Find the locus of R which divides QP externally in the ratio ½ : 1 .

At any point P on the parabola y2 – 2y – 4x + 5 =0, a tangent is drawn which meets the directrix at Q . Find the locus of R which divides QP externally in the ratio½ : 1 .

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At any point P on the parabola y² – 2y – 4x + 5 =0, a tangent is drawn which meets the directrix at Q . Find the locus of R which divides QP externally in the ratio
½ : 1 .