Question

Find an equation of the tangent line to the ellipse: x^2/a^2+ y^2/b^2 = 1 at the point (x0,y0)?

Answer 1

Do you already have your tangent line?

Answer 2

Yeah
The tangent line is supposed to be (x0x)/a^2 + y0y/b^2

Answer 3

I can't seem to simplify my answer to get that.

Answer 4

=1

Answer 5

Wasn't that supposed to be the ellipse?

Answer 6

Nope. The ellipse is x^2/a^2+ y^2/b^2 = 1. It's really similar though. I'm kinda wondering if my textbook messed up or something.

Answer 7

Are you in Calc AB? Or BC?

Answer 8

Uh... I'm in IB Math HL, so I'm not sure. What I'm doing is right after Pre-Calc though.

Answer 9

Ahhh, I see. I'm in Calc AB, but I think what my teacher told me might help. The slope of a line (like, say, maybe a tangent line) is -A/B. What's really confusing me about this are the variables that are being used. I'm used to working with the actual numbers instead of a and b.

Answer 10

Yeah, I know what you mean. I breezed through the number questions, but I just got a bit stuck here.

Answer 11

You might also want to take the derivative of the actual ellipse (though, I'm not sure if they teach that in IB Math; I'm thinking they do?). Then you would find out with the derivative of the ellips is equal to the tangent line's slope (-A/B).

Answer 12

Yeah, I took the derivative then got\[{2b ^{3}x(a-a'x) + 2ya ^{3}(y'b-b'y)}\div(a ^{3}b ^{3})\]

Answer 13

I'm trying to simplify it by inserting it into y-y1 = m(x-x1), but it isn't goign so well.

Answer 14

The entire equation should be equal to 0 (the one you just wrote out), and wherever you have y', there is also a dy/dx (the thing you're after). Once you get dy/dx by itself, that's the derivative. Then you find where that equals the slope of your ellipse at -A/B, and your tangent line would be (y-y1)=(dy/dx)(x-x1).

Answer 15

Ah, I hate expanding. I'll see if that'll work, thanks either way!

Answer 16

Oh, and a and b should be arbitrary, so their derivatives should be zero.

Answer 17

oh right I forgot that. that should help things.

Answer 18

so treating a and b as constants instead of variables got me a lot father: I ended up with \[y0(y-y0)\div b ^{2}=x0(x0-x)\div a ^{2}\]

Answer 19

You think there's any way I can get a 1 out of that mess?

Answer 20

You don't need a 1, right? 1 is just what the original ellipse is equal to. the derivative is the slope at a specific point. Where's your dy/dx?

Answer 21

The y' is \[(-xb^{2})\div (ya^{2})\]

Answer 22

Alright, and since, at the end of the question, you were asked to find he tangent line (the line of the slope) at the point (x0,y0), you would first plug x0 and y0 into your y'. What y' ends up equaling is the m in your equation (y-y1)=m(x-x1). And x1 and y1 are x0 and y0. That would be your tangent line.

Answer 23

yup. that's how I got y0(y−y0)÷b2=x0(x0−x)÷a2

Answer 24

no wait I think I got it

Answer 25

If I use the original equation, x0^2/a^2 = 1- y0^2/b^2 right?

Answer 26

So if I plug that into the equation I got above... the two y0^2/b^2 cancel and I'm left with 1!

Answer 27

=) Is your questioned answered now?

Answer 28

yup thanks :D

Answer 29

No problem, glad I could be of help ^-^.

Answer 30

Wish I could help you out but your question's way beyond my ability. if no one answers here you could try Yahoo Answers. There's some pretty smart people on there sometimes.

Answer 31

Alright, and don't sweat it ;). I'll go ahead and try that now since It's getting close to 9:30 and I've still got to finish my paper.

Answer 32

Good luck!