## Yahoo! 2 years ago 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 ?

1. Yahoo!

@shubhamsrg

2. Yahoo!

dy/dx =x^2/(6y)

3. Yahoo!

4. Yahoo!

Nw wat to do ?

5. shubhamsrg

-6y/x^2 is the slope of a line which cuts of equal intercepts as x axis.

6. shubhamsrg

hence slope should be equal to 1 or -1.

7. shubhamsrg

"-6y/x^2 is the slope of a line which cuts of equal intercepts ON THE axes.

8. shubhamsrg

|dw:1359206768852:dw|

9. Yahoo!

Did nt get u

10. shubhamsrg

-6y/x^2 is the slope of the normal right ?

11. Yahoo!

yup

12. shubhamsrg

and we are given that the normal cuts of equal intercepts on the axes.

13. Yahoo!

ok...then

14. Yahoo!

let the Points be (a,0) and (0,a ) ryt ?

15. 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.

16. Yahoo!

(a-0) / (0-a) = -1

17. shubhamsrg

What I mean is -6y /x^2 = 1 and/or -1

18. Yahoo!

ok...

19. shubhamsrg

we also know y = x^(3/2) /3 or -x^(3/2) /3

20. Yahoo!

yes

21. shubhamsrg

substitute in the previous eqn to find the required value(s) of x

22. Yahoo!

Yeah got it..:) thxx

23. shubhamsrg