NIMCET 2024 Paper

Section: Mathematics

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Question : 24 of 120

Marks: +1, -0

The two parabolas

y^2=4 a(x+c)

and

y^2=4 b x, a>b>0

cannot have a common normal unless:

$c>2(a+b)$
$c>2(a-b)$
$c< 2(a-b)$
$c< \;{2}/{a-b}$

Solution:

Equation of normal of the parabola

y^2=4 a(x+c)

is

\;y=m(x+c)-2 a m-a m^3

\;⇒ y=m x+m c-2 a m-a m^3

Similarly, equation of normal for the parabola

y^2=4 b x

is:

y=m x-2 b m-b m^3

As the normal is common, hence

\;m c-2 a m-a m^3=-2 b m-b m^3

\;(a-b) m^3=m(c+2 b-2 a)

\;⇒ m^2=\;{c+2 b-2 a}/{a-b}=\;{c}/{a-b}-2

As

m^2>0

, hence

\;{c}/{a-b}-2>0 ⇒ c>2(a-b)

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