Question

The points on the curve 9y^2=x^3 where the normal to the curve makes equal intercepts with the axes=?

Answer 1

@hartnn

Answer 2

did u differentiate to get dy/dx ?

Answer 3

yes

Answer 4

\[\LARGE 18yy'=3x^2=>y'=\frac{3x^2}{18y}=\frac{x^2}{6y}\]

Answer 5

the slope of the normal will be
\(\Large \dfrac{-1}{\dfrac{dy}{dx}}\)

Answer 6

normal slope..
\[\LARGE \frac{-6y}{x^2}\]

Answer 7

not getting how to "apply" the 2nd part of the question

Answer 8

let the point be x1, y1

make an equation of line since you have point and slope

Answer 9

then find out the intercepts, equate them

Answer 10

you know how to find intercepts, right ?

Answer 11

\[\large y-y_1 =\frac{-6y}{x^2} (x-x_1)\]
umm put x=0 to get y intercept and similarly for x?

Answer 12

exactly! :)

Answer 13

then ill get (0,0) ?

Answer 14

both intercepts 0? :/ how is that possible

Answer 15

ok, wait,
i think what we need to do is,
express that line equation in the form of
x/a +y/b=1

then a and b are intercepts

Answer 16

this is a curve but :O

Answer 17

oh okay

Answer 18

equating a and b , you will get one equation.

the other equation is the fact that x1,y1 lies on the curve
9y1^2 = x1^3

Answer 19

hogaya :D thanks!

Answer 20

welcome ^_^