How the FUCK do you find a tangent line of an ellipse [math]

>>8890479
algebraicly or geometricly?

>>8890479
apply an affine transformation that turns it into a circle, draw the tangent line to the circle at the corresponding point, then apply the inverse transformation.

>>8890483
I need to derive a formula [math] \frac{xx1}{a^{2}} + \frac{yy1}{b^{2}} = 1 [/math] (i don't know how to write '1' index in latex) from the formula of an ellipse given in OP, and the point T(x1, y1). I don't know what's that method called. Eurofag here

>>8890495
[math]x_{12_{123}345}[/math]

x_{123_{123}45}, one of the basics of TeX.

unrelated,
Why is TeX not working? wtf

[math] \lambda [/math]

you have to write an equation of all lines that intersect the ellipse at the point (x1,y1) then from that derive a tangent line that only intersects that one point in the ellipse.

>>8890511
Tex not werk bcoz mod is faget

>>8890511
Works for me. But wasn't mathjax CDN rip or something?

>>8890479
Anon, stop being stupid. The ellipse is a conic section, right? So, if you substitute a linear equation into x^2 or y^2, then you will get a quadratic expression of one variable. Pick your linear equation to be a possible tangent line with an unknown slope, but programmed such that the line will pass through whatever point on the ellipse you are interested in. Now, the equation formed by the composition of the ellipse equation and the linear equation, of one variable, will only be solved if and only if that variable belongs to some coordinate (x,y) such that the coordinate belongs to the ellipse and the linear equation (of course, one equation isn't enough to specify all of the points (x,y), you need the linear equation to compute the other variable based on the first variable the composite equation gave you). Now, the composite equation can be easily solved with the quadratic formula, but the slope of the linear equation is still unknown. But, you know that a tangent line of an ellipse should only intersect the ellipse once, as ellipses don't curve around much, so the result of the quadratic formula should be a single root. Using that constraint, you can figure out the slope of the linear function. And you programmed the linear function to include the point of the ellipse you were interested in, so now you have a tangent line of the ellipse.

did you do it?

>>8890522
I did that. Got a huge cubic equation for slope. I can't calculate this shit this way, it's so fucking brute force.

>>8890566
No, I did not. I hate my life.

>>8890572
then maybe you should accept calculus results.

>>8890572
Dumbass, it is not supposed to be cubic. Take y = m*x + b. Now, substitute y into the ellipse equation. The resulting equation will be of one variable, x, and have degree 2.

Brainlet here, maybe you could take the limit of p1-p2 where p1 is the point you want to calculate the tangent line of, and p2 approaches p1

Though that is quite similar to the calculus approach

>>8890577
when you make delta equal zero (1-root constraint) you end up in a quartic equation, no?

File: asdf.png (188KB, 589x364px) Image search: [Google]

188KB, 589x364px

>>8890574
I do accept it, however my Analytic Geometry textbook doesn't.

pic related

>>8890577
Yeah, and to get the tangent line I need to have only one solution to the equation you're talking about (cuz it's basically equation of points of intersection - and the tangent like only has one). I need to find the fucking 'm' and 'b', and when I try to do that I get a huge fucking equation.

>>8890580
That's literally the calculus approach.
I can't use fucking limits or anything not related to Analytic Geometry.

>>8890602
>>8890616
No, when you set the discriminant equal to 0, you don't use 0 = sqrt delta to find the slope, you first use the quadratic formula, but with the assumption that the discriminant is 0. This keeps the resulting equation in degree 2, which you then can use the quadratic formula a second time to get the slope.

File: wtf.jpg (105KB, 1440x900px) Image search: [Google]

105KB, 1440x900px

>>8890652
...you're right.
I fucking solved it.

I feel so retarded right now. Thank you wise anon

>>8890667
no prob, i just read everything off of this
http://www.ms.uky.edu/~pkoester/teaching/Old/Calc1_Spring2009//Notes/AlgTang.pdf

>>8890479

Let A = (-a, 0), B = (0, b), C = (a, 0), and D = (0, -b). Let P be the desired tangent point.

Let U be the intersection of DC and PB and X be the intersection of DB and PC.

Let V be the intersection of DA and PC and Y be the intersection of DC and PA.

Let T be the intersection of UX and VY.

Then TP is the desired tangent.