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# Find the locus  foot of perpendicular from focus to the tangent y^2=4ax

5 years 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)+ 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.

5 months ago
Tangent to parabola is, y = mx +a/m ....(i)
A line perpendicular to tangent and passing from focus (a, 0) is, y =-x/m +a/m ......(ii)
Solving both lines (i) and (ii) ⇒ x = 0