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The foot of the perpendicular from the focus on any tangent to a parabola lies on the tangent at the vertex.
Equation of the perpendicular to the tangent ty = x + at2Β β¦ (1)
From the focus (a, 0) is tx + y = at. Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β β¦ (2)
By adding (1) and (2) we get x = 0.(Since (1 + t2)Β β 0)
Hence, the point of intersection of (1) and (2) lies on x = Β 0 i.e. on y-axis. This is also the tangent at the vertex of the parabola.
The tangent at any point P on aΒ parabolaΒ bisects the angle between the focal chord through P and the perpendicular from P on the directrix.
?The tangent at P (at2, 2at) is ty = x + at2.
It meets the x-axis at T(βat2, 0).
Hence, from the figure given aboveΒ ST = SA + AT = a (1 + t2).Β
Also, SP = β(a2(1 + t2)2Β + 4a2Β t2Β ) = a(1 + t2) = ST, so thatΒ
β MPT = β PTS = β SPT β TP bisects β SPM.
The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus.
Let P(at2, 2a), be a point on the parabola y2Β = 4ax.
Then the equation of tangent at P is ty = x + at2.
Point of intersection of the tangent with the directrix x + a = 0 is (βa, at β a/t).
Now, slope of SP is (2at-0)/(at2-a) = 2t/(t2-1)Β and slope of SK is (at-a/t-0)/(-a-a) = -(t2-1)/2tΒ
β (Slope of the SP).(Slope of SK) = β1.
Hence SP is perpendicular to SK i.e. β KSP = 90Β°.Β
Tangents at the extremities of any focal chord intersect at right angles on the directrix.Β
?Let P(at2, 2at) and P(at12, 2at1) be the end points of a focal part on the parabola. Then t.t1Β = β1. Equations of the tangents at the point P and the point Pβ are ty = x + at2Β and t1y = x + at12Β respectively.
Let these tangents intersects at a point (h, k). Then h = att1Β and k = a(t + t1).
Since the tangents are perpendicular, tt1Β = β 1 β h β a.
Hence the locus of the point (h, k) is x = βa which is the equation of the directrix.Β
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