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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
 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
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
6y/x^2 is the slope of a line which cuts of equal intercepts as x axis.
 one year ago

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

shubhamsrgBest 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.
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
6y/x^2 is the slope of the normal right ?
 one year ago

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

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

shubhamsrgBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

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