I need a step by step tutorial on this basically i have no background on how to solve this. Would really appreciate the help.
Find the equation of the tangent line and normal line to the curve 2x^2+y^2=4 at (1, -sqrt(2)).

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- wolf1728

2x^2+y^2=4
Wouldn't you first have to put this into the form y = ???

- anonymous

I have to isolate y first?

- wolf1728

Okay isolating y
y^2=-2x^2 +4
y = Square root (-2x^2 +4 )

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- anonymous

then what?

- wolf1728

valtajaros
When someone asks a question then just leaves, I am not sticking around

- wolf1728

It helps if the asker sticks around so that I can ask some things

- anonymous

the equation for a line passing through a point can be written as \[y-y _{1}=(x-x _{1})m\]
where m is the slope of the line and \[y _{1}\] and \[x _{1}\] are the coordinates of the point that are known to us , in this case (1,-sqrt(2))
since the point lies both on the curve and on the line, the slope at this point will be same for curve and the line.
we know slope=dy/dx
derive the given equation and put the values of x and y as, 1 and sqrt(2) respectively, you get the value of m
put the values of m, x, and y in the general equation and you get the equation for tangent

- wolf1728

Okay, nitishdua31 seems aware of this so I will leave.

- anonymous

really sorry im having internet problems thank you tho

- anonymous

so first i have to get the derivative of y=sqrt(4-2x^2)

- anonymous

http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-tannorm-2009-1.pdf
see if this can help :)

- anonymous

thanks man ill have a look in to this! :)

- anonymous

yes find the derivative and that gives you the value of slope

- anonymous

happy to help :D

- anonymous

so like sqrt(u) = du/2sqrt(u) from y = sqrt(4-2x^2)
i get y' =(-4x/2sqrt(4-2x^2) = -2/sqrt(4-2x^2)

- anonymous

*-2x/sqrt(4-2x^2)

- anonymous

-2(1)/sqrt(4-2(1)^2) = -2/sqrt(2) ???

- anonymous

yes ! so its basically -sqrt(2)

- anonymous

Attachment from Mathematica 9 may be a bit short on comments.

##### 1 Attachment

- wolf1728

valtajaros
"really sorry im having internet problems thank you tho"
**************************************************************
sorry :-(

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