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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 + at2 … (1)
From the focus (a, 0) is tx + y = at. … (2)
By adding (1) and (2) we get x = 0.(Since (1 + t2) ≠ 0)
Hence, the point of intersection of (1) and (2) lies on x = 0 i.e. on y-axis. This is also the tangent at the vertex of the parabola.
The tangent at any point P on a parabola 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(–at2, 0).
Hence, from the figure given above ST = SA + AT = a (1 + t2).
Also, SP = √(a2(1 + t2)2 + 4a2 t2 ) = a(1 + t2) = ST, so that
∠MPT = ∠PTS = ∠SPT ⇒ TP bisects ∠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 y2 = 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 (–a, at – a/t).
Now, slope of SP is (2at-0)/(at2-a) = 2t/(t2-1) and slope of SK is (at-a/t-0)/(-a-a) = -(t2-1)/2t
⇒ (Slope of the SP).(Slope of SK) = –1.
Hence SP is perpendicular to SK i.e. ∠KSP = 90°.
Tangents at the extremities of any focal chord intersect at right angles on the directrix.
?Let P(at2, 2at) and P(at12, 2at1) be the end points of a focal part on the parabola. Then t.t1 = –1. Equations of the tangents at the point P and the point P’ are ty = x + at2 and t1y = x + at12 respectively.
Let these tangents intersects at a point (h, k). Then h = att1 and k = a(t + t1).
Since the tangents are perpendicular, tt1 = – 1 ⇒ h – a.
Hence the locus of the point (h, k) is x = –a which is the equation of the directrix.
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