Question

find the tangent of the ellipse
\(\large \frac{x^2}{4}+y^2=1\)
at the point \(p (\sqrt 2, \frac{1}{\sqrt2})\)

Answer 1

@ganeshie8 @phi @jdoe0001 @myininaya

Answer 2

did you find y'?

Answer 3

\(\large y=\sqrt (1-\frac{x^2}{x}) \)

Answer 4

why didn't you use implicit differentiation?

Answer 5

idk, u told me "did you find y'? "
so i thought of finding y :O

Answer 6

do you know how to do implicit differentiation?
you can solve for y and find y' explicitly..
it just looks uglier.

Answer 7

could u plz show it ?

Answer 8

am I wrong by thinking -> \(\bf \cfrac{x^2}{4}+y^2=1\implies x^2+4y^2-4=0\quad ?\)

Answer 9

hmmm I see... nevermind that =)

Answer 10

Here is example to find y':


\[x^2+xy^2=1+yx \]

Take derivative of both sides with respect to x

\[\frac{d}{dx}(x^2+xy^2)=\frac{d}{dx}(1+yx)\]

Use the sum rule for differentiation
\[\frac{d}{dx}(x^2)+\frac{d}{dx}(xy^2)=\frac{d}{dx}(1)+\frac{d}{dx}(yx)\]

We know how to find derivative of x^2, it is 2x

now for finding derivative of (xy^2) or (yx) we need to know the product rule since we have a product
\[2x+y^2 \frac{d}{dx}(x)+x \frac{d}{dx}(y^2)=\frac{d}{dx}(1)+y \frac{d}{dx}(x)+x \frac{d}{dx}y\]

now I'm going to go ahead and replace the derivative of 1 with 0
since we know derivative of any constant is 0

now for find derivative of y^2 or y we need to know chain rule

(y^2)'=2y*y' <---remember y=y(x)
(y)'=y'
so we have now
\[2x+y^2+2yy'=y+xy'\]
Then you can solve this pretty easily for y.
By putting your terms that have a y' together and put everything that doesn't have a y' on the other side of the equation like this:

\[2yy'-xy'=y-2x-y^2 \]
We do this so we can factor out the y' and divide both sides by what y' is being multiplied by to obtain what y' is
like so:
\[y'(2y-x)=y-2x-y^2 => y'=\frac{y-2x-y^2}{2y-x}\]

Answer 11

btw , i was thought of
r(t)=[2 cos t , sin t ]
i sow it in the text book but dnt know what it is :\
then it show
r'(t)=[- sin t, cos t)

Answer 12

thax @myininaya its awesome !!

Answer 13

ok you try differentiating x^2/4+y^2=1 to find y' now

Answer 14

\[\frac{d}{dx}(\frac{x^2}{4})=?\]

Answer 15

1/2 x ?

Answer 16

ok
good now
\[\frac{d}{dx}(y^2)=?\]

Answer 17

I actually already told you this one..

Answer 18

yeah !
im still reading ur sol !
its a bit long , but thx for detail :)
i cant understand it nw
thank you :D

Answer 19

ok well do you know how to find derivative of (4x-1)^2?

Answer 20

8(4x-1)

Answer 21

and you did that by finding derivative of the inside times the the derivative of the outside
so like you did this:
2(4x-1)*(4)

Answer 22

well what if y=4x-1 and y'=4
so we have derivative of y^2 is 2(y)*y'
right?

Answer 23

right

Answer 24

so when we differentiate x^2/4+y^2=1
what do we get as the very next step

Answer 25

y2=1- x^2/4
2yy'=-x/4
?

Answer 26

that is actually really really good
now
solve for y'

Answer 27

oops and that is 2yy'=-x/2 right?

Answer 28

y'=x/4y

Answer 29

yeah u right its 2 not 4

Answer 30

and we have y'=-x/(4y)

Answer 31

great

Answer 32

now we want to find the tangent at the point \[(\sqrt{2},\frac{1}{\sqrt{2}})\]
So you found the formula for the general slope of our ellipse

Answer 33

plug this point into find our slope m

Answer 34

yeah so now
if we put x,y values
its clear that
y' at p=- 4

Answer 35

then we could just use point slope form to find the line
\[y-y_1=m(x-x_1)\]

Answer 36

cool !
i got it ur great ;)

Answer 37

wait
i'm not getting -4

Answer 38

\[y'|_{(\sqrt{2},\frac{1}{\sqrt{2}})}=-\frac{\sqrt{2}}{4 \cdot \frac{1}{\sqrt{2}}}\]
is this what you did

Answer 39

yess

Answer 40

ohh its -1/2 then ?

Answer 41

\[=-\frac{\sqrt{2}}{4} \cdot \frac{1}{\frac{1}{\sqrt{2}}}=-\frac{\sqrt{2}}{4} \cdot \sqrt{2}\]
right

Answer 42

now you got it

Answer 43

thanks again !

Answer 44

np

Answer 45

If you haven't learned implicit differentiation, that will be around the corner. Have fun.

Answer 46

huh no i learned it , im just refreshing rusty concepts :)

Answer 47

ah