Question

Any one please help me with this question or tell me how to go about it ??

Find the interval for "s" so that (3-s)x + sy + (s^2-1)=0 is normal to the curve xy = 4 .

Answer 1

@terenzreignz

Answer 2

@zepdrix

Answer 3

hmm, orthogonal functions come to mind

Answer 4

\[\LARGE y'=\frac{-(3+s)}{s}\] after differentiating the equation
orthogonal functions o.O

Answer 5

start with derivatives; if the derivatives are perpendicular then they will be orthogonal

Answer 6

\[\LARGE y'=\frac{-y}{x}=\frac{x}{y}\]

Answer 7

Now i am supposed to equate both but i dont get anything?

Answer 8

(3-s)x + sy + (s^2-1) = 0

(3-s) + s y' = 0; y' = (s-3)/s
----------------------

xy = 4

y + xy' = 0

y' = -y/x = -4/x^2

Answer 9

\[\frac{s-3}{s}=-\frac{1}{-4/x^2}\]
\[\frac{4}{x^2}(s-3)=s\]
\[4(s-3)=sx^2\]
etc ...

Answer 10

so how do we get the interval??

Answer 11

notice the since s is in a denominator along the way the s=0 is bad for everyone, thats just an observation

Answer 12

did yours just goofy too?

Answer 13

\[4(s-3)=sx^2\]
\[4(1-\frac 3s)=x^2\]
\[4-\frac {12}s=x^2\]
notice that the range of x^2 is from 0 to +inf

Answer 14

yes it is

Answer 15

\[0\le4-\frac{12}s\le\infty\]
\[-4\le-\frac{12}s\le\infty\]
\[-\infty\le\frac{12}s\le4\]
getting any ideas yet?

Answer 16

\[\frac{12}s\le 4\]
\[\frac{12}4\le s\]