If focus of a parabola is (1, 2) and feet of perpendiculars on any two tangents to this parabola from focus are (3, 4) and (4, 6). Then vertex of the parabola is
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Asked on December 20, 2019 by
Sujatha Punjabi
The locus of the foot of the perpendiculars drawn from the vertex on a variable tangent to the parabola y
2
=4 ax is
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The foot of perpendicular from focus to the tangent y
2
=4ax equation of the perpendicular to the tangent is
ty=x+at
2
→ eqn (1)
So, from the focus (a,0) is tx+y=at→(2)
by adding eq (1) & eq (2) we get, fig a
x=0(as(1+t
2
)
=0)
∴ the point of intersection of (1) & (2)
lies on x=0 i.e. on −y axis and this is the tangent at the vertex of the parabola.
As the tangent at the point (P) on a parabola bisects the angle between the focal chord through P and perpendicular from P on the directrix.
∴ the tangent at p(at
2
,2at) is ty=x+at
2
meets x− axis at T(−at
2
,0)
Hence, from the figure we can analyse that
ST=SA+AT=a(1+t
2
)
also SP=
a
2
(i+t
2
)+4a
2
t
2
=a(1+t
2
)=ST
which means ∠MPT=∠PTS=∠SPT=TP
Let p(at
2
,2a) be a point on the parabola y
2
=4ax
equation of the tangent P is ty=x+at
2
Point of intersection of the tangent with the directrix is ratio where (−a,(at−a)/t)
Now, slope of SP is (2at−0)/(at
2
−a)=2t/(t
2
−1)
parallel slope of SK is (at−at/0)/(−a−a)=−(t
2
−1)/2t
⇒ (Slope of the SP) (Slope of SK)=−1 there by SP is perpendicular to SK i.e., ∠KSP=90
o
as we know that Tangent at teh extremities of any focal chord intersect at right angels on the directrix.
is p(at
2
,2at),p(at
1
2
,2at
1
) are end points we get t.t
1
=−1⇔ eq of tangent ty=x+at
2
& t
1
y=x+at
1
2
if these tangent intersect ar a point (h,k) then
h=att
1
and k=a(t+t
1
).
∴ the tangent are perpendicular , tt
1
=−1⇒h−a