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The x co-ordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?

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dy/dx =x^2/(6y)
gradient of normal is -6y/(x^2)

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Other answers:

Nw wat to do ?
-6y/x^2 is the slope of a line which cuts of equal intercepts as x axis.
hence slope should be equal to 1 or -1.
"-6y/x^2 is the slope of a line which cuts of equal intercepts ON THE axes.
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Did nt get u
-6y/x^2 is the slope of the normal right ?
yup
and we are given that the normal cuts of equal intercepts on the axes.
ok...then
let the Points be (a,0) and (0,a ) ryt ?
any line which cuts off equal intercepts on the axes must have a slope of 1 or -1 right? Since it is making an angle of pi/4 or 3pi/4 with the +ve X axes.
(a-0) / (0-a) = -1
What I mean is -6y /x^2 = 1 and/or -1
ok...
we also know y = x^(3/2) /3 or -x^(3/2) /3
yes
substitute in the previous eqn to find the required value(s) of x
Yeah got it..:) thxx
glad you did. ^_^

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