How the FUCK do you find a tangent line of an ellipse [math] frac{ x^{2} }{a^{2} } + frac{ y^{2} }{b^{2} } = 1[/math]...

How the FUCK do you find a tangent line of an ellipse [math] \frac{ x^{2} }{a^{2} } + \frac{ y^{2} }{b^{2} } = 1[/math] in a point WITHOUT using calculus???

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ms.uky.edu/~pkoester/teaching/Old/Calc1_Spring2009//Notes/AlgTang.pdf
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algebraicly or geometricly?

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.

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

[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.

Tex not werk bcoz mod is faget

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

user, 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.