Question

Hey ! What is the eqn of the tangent line ((x-2)^2)/4+(y+1)^2=1 @ point (3,-(2+sqrt3)/2)? i think the derivative of the eqn is (-1/2x+1)/2y+2.Helpp:))

Answer 1

is this \[\frac{(x-2)^2}{4+(y+1)^2}=1\]?

Answer 2

if so first step would be to write
\[(x-2)^2=4+(y+1)^2\]

Answer 3

or maybe even
\[(x-2)^2-(y+1)^2=4\]

Answer 4

noo..almost..the 1st one was more right..excep tht the (y+1)^2 was added

Answer 5

Satellite:its an ellipse :P That'll show you how to fix the equation.

Answer 6

\[(x-2)^{2}/4 + (y+1)^{2}\]

Answer 7

Satellite:its an ellipse :P That'll show you how to fix the equation.

Answer 8

ok let me try again.
\[\frac{(x-2)^2}{4}+(y+1)^2=1\]?

Answer 9

yess:))

Answer 10

ooh yes @malevolence got it i am a little slow. other one was hyperbola i think

Answer 11

Yeah haha

Answer 12

ok so taking derivatives just as easy we get
\[\frac{x-2}{2}+2(y+1)y'=0\]

Answer 13

lets see if i can solve this for y' without messing up. i get
\[y'=\frac{-x+2}{4(y+1)}\] hows that look?

Answer 14

good!:)

Answer 15

whew i messed up on the last one ok now we plug an chug yes?

Answer 16

i guess so..thts wer im stuck:/

Answer 17

oh well it looks pretty annoying.
\[x=3\]
\[y=-\frac{2+\sqrt{3}}{2}\]?

Answer 18

maybe not so bad. numerator of our fraction is 5 yes?

Answer 19

ok it is -1 fine

Answer 20

numerator is -1. now lets look at
\[y=-\frac{2+\sqrt{3}}{2}=-1-\frac{\sqrt{3}}{2}\]

Answer 21

first we add 1 to get
\[y+1=-\frac{\sqrt{3}}{2}\]

Answer 22

the - gets dsitributed in the 2+ sqrt 3..so the y coordinate is (-2-sqrt3 )/2 right

Answer 23

then we multiply by 4 to get
\[-2\sqrt{3}\]

Answer 24

yes if it is the way you wrote it we distribute the - sign

Answer 25

yeah:)

Answer 26

so after all is said and one, after we add 1, multiply by 4 we get the denominator is just
\[-2\sqrt{3}\] and so the whole fraction is
\[\frac{1}{2\sqrt{3}}\]

Answer 27

assuming i did not make another arithmetic mistake

Answer 28

@malevolence how'd i do?

Answer 29

i got..\[-1\div -4-2\sqrt{3}\]

Answer 30

if i plug in the x and y coordiante given

Answer 31

course you are not done yet. you have to find the equation of the line. use the almighty point-slope formula to get
\[y+\frac{2+\sqrt{3}}{2}=\frac{1}{2\sqrt{3}}(x-3)\] etc

Answer 32

@sanhita that is not what i got

Answer 33

sry..minez was wrong..

Answer 34

\[y=-\frac{2+\sqrt{3}}{2}\] yes

Answer 35

i actually got -1/2-sqrt 3

Answer 36

lets take it one step at a time

Answer 37

y=(-2-sqrt 3)/2...when we distribute the -

Answer 38

which is also
\[-1-\frac{\sqrt{3}}{2}\]

Answer 39

yes

Answer 40

now we add 1 yes?

Answer 41

denominator is
\[4(y+1)\]

Answer 42

yes..n the numerator wud b -2-sqrt 3?

Answer 43

first we need to add 1. then multiply by 4

Answer 44

yea..aftr doin tht...thts wat i got

Answer 45

sry not -2..2-sqrt 3

Answer 46

ok one step at a time
\[y=-1-\frac{\sqrt{3}}{2}\]

Answer 47

\[y+1=-1-\frac{\sqrt{3}}{2}+1=-\frac{\sqrt{3}}{2}\]

Answer 48

yup:)

Answer 49

and finally
\[4(y+1)=-2\sqrt{3}\]

Answer 50

ohhyeaa..rightt

Answer 51

and finally finally the whole fraction is
\[\frac{1}{2\sqrt{3}}\]

Answer 52

haha yes..sry abt the confusion:)

Answer 53

and finally finally finally you have to find the equation of the line. i started it somewhere above, i will get you finish it

Answer 54

well this is wer i am at so far...y+1+sqrt3/2=(1/(2*sqrt3))x-3/(2*sqrt3)

Answer 55

simplified it is...\[y=1x/\left( 2\sqrt{3} \right)-3/(2\sqrt{3})-1\]

Answer 56

thnks u soo much 4 ur work n patience!:)

Answer 57

looks like you got it. i am going to assume you did that last bit of arithmetic correctly. good work!