Question

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 ?

Answer 1

@shubhamsrg

Answer 2

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

Answer 3

gradient of normal is -6y/(x^2)

Answer 4

Nw wat to do ?

Answer 5

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

Answer 6

hence slope should be equal to 1 or -1.

Answer 7

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

Answer 8

Did nt get u

Answer 9

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

Answer 10

yup

Answer 11

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

Answer 12

ok...then

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

What I mean is

-6y /x^2 = 1 and/or -1

Answer 17

ok...

Answer 18

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

Answer 19

yes

Answer 20

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

Answer 21

Yeah got it..:) thxx

Answer 22

glad you did. ^_^