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Tangents drawn from point K to a parabola meets at P and Q. If S is the focus of the parabola, prove that SK^2= SP.SQ

Tangents drawn from point K to a parabola meets at P and Q. If S is the focus of the parabola, prove that SK^2= SP.SQ

Grade:11

1 Answers

Arun
25763 Points
one year ago
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°. 

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