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 2 years 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 ?
 2 years 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!
 2 years ago
Best ResponseYou've already chosen the best response.0gradient of normal is 6y/(x^2)

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

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

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

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

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

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

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

shubhamsrg
 2 years 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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