anonymous
  • anonymous
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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  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@shubhamsrg
anonymous
  • anonymous
dy/dx =x^2/(6y)
anonymous
  • anonymous
gradient of normal is -6y/(x^2)

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anonymous
  • anonymous
Nw wat to do ?
shubhamsrg
  • shubhamsrg
-6y/x^2 is the slope of a line which cuts of equal intercepts as x axis.
shubhamsrg
  • shubhamsrg
hence slope should be equal to 1 or -1.
shubhamsrg
  • shubhamsrg
"-6y/x^2 is the slope of a line which cuts of equal intercepts ON THE axes.
shubhamsrg
  • shubhamsrg
|dw:1359206768852:dw|
anonymous
  • anonymous
Did nt get u
shubhamsrg
  • shubhamsrg
-6y/x^2 is the slope of the normal right ?
anonymous
  • anonymous
yup
shubhamsrg
  • shubhamsrg
and we are given that the normal cuts of equal intercepts on the axes.
anonymous
  • anonymous
ok...then
anonymous
  • anonymous
let the Points be (a,0) and (0,a ) ryt ?
shubhamsrg
  • shubhamsrg
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.
anonymous
  • anonymous
(a-0) / (0-a) = -1
shubhamsrg
  • shubhamsrg
What I mean is -6y /x^2 = 1 and/or -1
anonymous
  • anonymous
ok...
shubhamsrg
  • shubhamsrg
we also know y = x^(3/2) /3 or -x^(3/2) /3
anonymous
  • anonymous
yes
shubhamsrg
  • shubhamsrg
substitute in the previous eqn to find the required value(s) of x
anonymous
  • anonymous
Yeah got it..:) thxx
shubhamsrg
  • shubhamsrg
glad you did. ^_^

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