Question

find the lines that are tangent and normal to the curve at the given point
x^2y^2=81 at the point (1,9)

Answer 1

to get the tangent line you have to find its slope and a point
now you have the point (1,9)
how to get the slope...any ideas ?

Answer 2

solve for x im guessing

Answer 3

look
The first derivative means the slope of the tangent

Answer 4

would the derivative be 2x2y

Answer 5

\[x ^{2}*y ^{2}=81\]
right form ?

Answer 6

ya

Answer 7

ok
how the result is 2x*2y ?

Answer 8

idk what to do

Answer 9

\[\frac{ d }{ dx }(x ^{2}y ^{2})=\frac{ d }{ dx }(81)\]

Answer 10

\[2x*y ^{2}+2yx ^{2}*\frac{ dy }{ dx }=0\]

Answer 11

\[\frac{ dy }{ dx }=\frac{ -y }{ x }\]

Answer 12

then what?

Answer 13

do you plug the point values in?

Answer 14

yes go ahead

Answer 15

dy/dx=?

Answer 16

-9

Answer 17

good
so the slope is -9
and the point is (1,9)
now can you create the tangent line ?

Answer 18

y=-9x+18

Answer 19

how would you find the line normal to the curve

Answer 20

the line normal to the curve at a point is also normal to the tangent line at the same point
so the slope of the normal line=-1/m
where m=slope of tangent line
so the slope of the normal line =??

Answer 21

so just the inverse which is -1/9

Answer 22

no 1/9 not -1/9

Answer 23

what would the equation be i got y=1/9x+82/9

Answer 24

80/9 not 82/9

Answer 25

cool thanks so much

Answer 26

wait
check this :)
the blue is your function
the red line is the tangent
and the green is the normal
at the same point