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 one year ago
The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?
 one year ago
The x coordinate 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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Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.0gradient of normal is 6y/(x^2)

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.16y/x^2 is the slope of a line which cuts of equal intercepts as x axis.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1hence slope should be equal to 1 or 1.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1"6y/x^2 is the slope of a line which cuts of equal intercepts ON THE axes.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359206768852:dw

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.16y/x^2 is the slope of the normal right ?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1and we are given that the normal cuts of equal intercepts on the axes.

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.0let the Points be (a,0) and (0,a ) ryt ?

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1any 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.

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1What I mean is 6y /x^2 = 1 and/or 1

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1we also know y = x^(3/2) /3 or x^(3/2) /3

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.1substitute in the previous eqn to find the required value(s) of x
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