KVPY SB SX Exam 2011 Question Paper

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Question : 8 of 120

Marks: +1, -0

In the xy-plane, three distinct lines concur at a point (λ , 0). Further the lines

l_1,l_2,l_3

are normals to the parabola

y^2 = 6x

at the points

A = (x_1,y_1)

,

B = (x_2,y_2),C = (x_3,y_3)

respectively. Then we have

$λ< -5$
$λ > 3$
$- 5 < λ < - 3$
$0 < λ < 3$

Solution:

Any normal

y = mx - 2am - am^3

Here

a = 3/2

through (λ, 0)

0 = mλ- 2am - am^3

m = 0, λ = 2a + am^2

m^2=λ/a-2 > 0

λ > 2a⇒ λ > 3

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