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Determine the equation of the tangent to the curve at a given point. x^2+9y^2=37 (1,2)

Mathematics
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I've gotten the slope of -1/18, but I don't know if that's correct. And if so, how do I continue?
So, first, dy/dx=-x/dy. I then subbed in (1,2) and got -1/18.
Lol uhh so I cant give answers so ill just leave now but what youv done so far is correct :D

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Other answers:

I don't need the answer. I need to know what my process afterwards should be.
What's the derivative?
And Chad is in 11th grade, he won't be able to help you
1. Derivative is 2x+18y(dy/dx)=0. Right? 2. I figured that out already...
lol ink I can give him an answer ;D
x^2+9y^2=37 dy/dx 37=dy/dx x^2+dy/dx 9y^2 0=2x+18y dy/dx
No, you won't give him the answer he's looking for. THe answer he's looking for is an explanation, because, as everyone knows, it's the explanantion that matters in calc
I already have the answer in my textbook and potentially Wolfram Alpha. I need to know how to GET there though.
Normally at this point I try to isolate dy/dx for convenience
from conic section?
0=2x+18y dy/dx (-2x)/(18y)=dy/dx x/(-9y)=dy/dx
Yeah, I think I did that.It's -x/dy right? @ Shruti: I don't think so...
And, my answer agrees with your answer here.
Although my work here is moot, because i'm in 9th grade.
Ok. Thanks for the help so far.
Ok. How would I get to the answer of x+18y-37=0 from knowing that the slope is -1/18?
eh?
If you had the derivative and wanted to get the original equation?
I guess?
@shruti By conics did you mean http://www.purplemath.com/modules/circle3.htm
since this is an equation of an ellipse, y can be considered an implicit function of x right?
I guess. My notes only reveal the steps. This work is not from a textbook.
In fact, it's not even addressed in my textbook. That's strange.
it can be .... to derive implicitly, ignore the form of the variable and treat everything as tho it acted like you normally think of deriving x^2+9y^2=37 ; just derive it all, but dont toss out the derived bits 2x x' + 18y y' = 0 now lets consider this when we derive it with respect to x x' = dx/dx = 1 y' = dy/dx like normal so, lets solve it for y'
If you have the derivative, and want to get the original equation, you take the anti-derivative of it
If you're asking about taking anti-derivatives (indefinite integrals), and have not yet learned about rieman sums, don't worry about the question
Otherwise, I might have to look into it.
Y' is -x/dy, which, when we sub in (1,2) is -1/18 right?
yes tim, -1/18 is right
What would I do next then?
Well, you have a point, and a slope.
(1,2) -> point -1/18 -> slope
remember, y=mx+b
Would I sub those into y=mx+b?
you have slope, you have a point, put it in point slope form...
put in the point and the slope, and then you have b
then, put in y=mx+b, cept don't put in that point
This isn't a calculus problem, this is an algebra problem, :D
Sorry, what is point slope form? It's algebra? I must really have been "dumbing" down these years...
Ok.
There's multiple ways of writing the equation for a line.
This is point slope: y-y1=m(x-x1)
this is slope intecept: y=mx+b
This is standard form: ax+by=c
I normally work with slope intercept, because i'm more used to it.
However, point slope is the fastest way to go here.
y-y1=m(x-x1) (x1,y1)=(1,2) m=-1/18
The answer is in the form of ax+by=c I guess, but I am also more familiar with slope intercept.
Wait, I already have the slope... I know it is -1/18. But how do I get to x+18y-37?
y-1=(-1/18)(x-2) now, just isolate y, and distribute x
They have given you the point that the line must pass through, and you have found the slope,
Oh. I finally understand. Thank you very much. I will call out again if I need help.
Ok. :D
Wait, 1 and 2 in y-1=(-1/18)(x-2)/ Shouldn't it be y-2=(-1/18)(x-1)? Because (1,2)?
Yes
I got it confused myself. xD
Why doesn't my answer look like that provided on the sheet, which is x+18y-37?
Oh. I figured that out. Sorry.
Thank you all for putting forth the effort to assist me.
Or at the very least, looking over the question.
No problem :)
This is good premptive practice for me ;)

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