IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

casio_ss

Find the condition that parabolas y^2 = 4b(x - c) and y^2 = 4ax have a common normal other than the x-axis (a > 0 , b > 0 )...

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magiclko

i think the other parabola is y² = 4ax

the equation of a nromal to the parabola (y-k)²= 4a(x-a) is

y = mx -2am - am³

thrfore, here the normals will be

y = mx -2am - am³for parabola y² = 4ax

and

y = m(x-c) -2bm - bm³for parabola y² = 4b(x - c)

bothe the equations represent same line, thrfore

2am + am³= 2bm + bm³+ cm

=> 2(a-b) + m² (a-b) -c = 0

= > m² = c - 2(a-b)

(a-b)

which should be always gr8r than 0

so c - 2a + 2b > 0

casio_ss

sorry!!!!

the answer is

2a < 2b + c

and yes that is y^2 = 4ax

u hv not commited any mistake..

akhileshsk

y=m(x-c)-2bm-bm^3

y=mx-2am-am^3

equate the constant terms

and u get

m^2(a-b)+2(a-b)-c=o

c-2a+2b>o

thats the ans

rate it

magiclko

edited it, check it out