IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - Parabola-2

nishantsingh89

y = mx -2am - am^3 { for parabola y^2=4ax}

so here eqn of noormal bcomes

y= mx-8m-4m^3 ..........(1)

the other eqn is y = xcos(theta) -11cos(theta)-cos3(theta)
y= xcos(theta) - 11cos(theta)-(4cos^3 (theta) - 3cos(theta))

y = x cos(theta) - 8cos(theta) -4cos^3(theta) .......(2)

now compare 1&2

see m is slope = tanh { h be any angle}

now on
-1<=cos(theta)<=1
always
now frm 1 we see that slope is tanh, frm 2 is cos(theta)
since both eqn are identical
we must have tanh=cos(theta)
now see here that range of cos(theta) is restricted but not of tanh
so
whatever b the value of cos(theta) there will be a value of 'h'
such that
tanh=cos(theta)

hence the given equation will always be a normal
for any theta

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