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Question

Two mutually perpendicular tangents to the parabola y2=4ax meet the axes in P1 and P2 . If S is the focus of the parabola then 1l(SP1)+1l(SP2) is equal to

A
4a
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B
2a
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C
1a
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D
14a
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Solution

The correct option is C 1a
Let the tangents be at point

P(at21,2at1),QP(at22,2at2)

we know the point of intersection of tangents at any two given points

P(at21,2at1),Q(at22,2at2)is (at1.t2,2a(t1+t2))

since the tangents are mutually perpendicular and by property of tangents Mutually perpendicular tangents to a parabola meet on the directrixi.e

x=a hence at1.t2=aort1.t2=1

Eqn of tangent to a parabola

y=mx+am

where m is slope of tangent

y2=4ax

2ym=4a

m=2ay

for P,m1=2a2at1=1t1

y=m1x+am1

y=xt1+at1

P1=puty=0

x=at21

P1(at21,0)

for Q,m2=2a2at2=1t2

y=xt2+at2

P2=puty=0

x=at22

P2(at22,0)

L(SP1)=a(at21)=a(1+t21)

L(SP2)=a(at22)=a(1+t22)

from 1 above

t1.t2=1

1L(SP1)+1L(SP2)=1a(1+t21)+1a(1+t22)=1a⎜ ⎜11+t21+11+1t21⎟ ⎟=1a(1+t211+t22)=1a

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