Question

Find the equations of the tangent lines at the point the curve crosses itself:
x=t^3-6t y= t^2
I came up with the following

Answer 1

\[y= \pm \frac{ 12 }{ \sqrt{6} }x\]

Answer 2

what did you do?

Answer 3

\[t _{1}^{3}-6t _{1}=t _{2}^{3}-6t _{2}.....AND....t _{1}^{2}=t _{2}^{2}\]

Answer 4

correct but you also need to assume that \[t_1 \neq t_2\]

Answer 5

\[t _{1}neqt _{2}\]

Answer 6

I did left that out...So t1=-t2

Answer 7

and t2=-t1

Answer 8

you only need t1 = -t2. There is not need for the other equation since you just move the negative to the other side.

Answer 9

so the two values of the parameter is -sqrt(6) and sqrt(6) yes?

Answer 10

I substituted back into the first equation and ended up with \[2t _{1}(t _{1}^{2}-6)=0 ....thus t=0 ..and..t=\sqrt{6}\]

Answer 11

The lines crosses itself at (0,6)

Answer 12

if t1= 0, then t2 = 0 but we already assume t1 =/= t2.

also,
t^2 - 6 = 0
t^2 = 6
t =+/- sqrt(6)

Answer 13

\[\frac{ 2t }{ 3t ^{2}-6 }\]

Answer 14

or \[\frac{ 2\sqrt{6} }{ 12 }\]

Answer 15

that's one slope. What about the other one?

Answer 16

Negative

Answer 17

right. Btw, simply the value of slope further.

Answer 18

\[y=\frac{ \sqrt{6} }{ 6 }x-\sqrt{6}\]

Answer 19

and\[y= -\frac{ \sqrt{6} }{ 6 }x+\sqrt{6}\]

Answer 20

it's \[y=\frac{ \sqrt{6} }{ 6 }x+\sqrt{6}\]

Answer 21

the point is (0,6) so answer is
y = +/- sqrt(6)x/6 + sqrt(6)

Answer 22

\[\frac{ 2\sqrt{6} }{ 12 }(x-6)=\frac{ 2\sqrt{6} }{ 12 }-\frac{ 12\sqrt{6} }{ 12 }\]

Answer 23

again the point is (0,6). NOT (6,0)

Answer 24

My bad...

Answer 25

no prob

Answer 26

the sqrt (6) remains positive for both equations?

Answer 27

i made a mistake

it's y = +/- sqrt(6)/6x + 6

Answer 28

yes it's positive for both since 6 is the y-iintercept

Answer 29

I see thank you you've been a great help