Equation of tangent from a point outside it

Question

Is there any general method of finding the equation of the tangent of a function $f(x)$ from a point $(a,b)$? $\hspace{1 mm}$ Then how do you find the angle between two tangents from $(0,0)$ to a parabola $y^2 = 4a(x-a)$?

Answer 1

general method will depend on how complex the function $f$ is. for simple polynomials you can find the intersection of a line and the graph of $y = f$ and require the repeating roots for the line to be a tangent. if the function is complex, you do the following steps:

(a) you pick a point $(k, f(k)$

(b) get the tangent at this point $y - f(k) = (x - k)f^\prime(k)$

(c) see the condition required for the point to be on this on tangent. that will will determine the contact points $(k, f(k)$

let us apply this in the case of the point $(0,0)$ and the graph of $y^2 = 4a(x-a)$

first method:

any line through $(0,0)$ can be written as $y = mx$ and $m$ to be determined. substituting this in $y^2 = 4a(x-a),$ we get a quadratic equation $$m^2x^2 -4ax + 4a^2 = 0$$ for repeating roots, the discriminant $16a^2 - 16a^2m^2 = 0$

that gives you $$m = \pm 1, x = \dfrac{4a}{2m^2} = 2a, y = \pm 2a $$

second method:

pick a point $(a+ \dfrac{k^2}{4a}, k)$ on the graph of $y^2 = 4a(x-a)$ where $k$ will be determined later. the tangent at this point satisfies $$2y dy = 2k(y - k) = 4a(x - a - \dfrac{k^2}{4a}) = 4a dx$$ for the point $(0,0)$ to be on this line we need $$ 2k(0 - k) = 4a(0 - a - \dfrac{k^2}{4a})$$ which is equivalent to $$k^2 = 4a^2, k = \pm 2a $$ as before.

Answer 2

For sufficiently regular functions such as polynomial or rational functions, you can take a straight line with a variable slope $t$ passing though $(a,b)$ and determine its points of intersection with the curve $y=f(x)$. You have a tangent line if the resulting equation has a double root.

In the example you give, you obtain the equation $t^2x^2=4a(x-a)\iff t^2x^2-4ax+4a^2=0$. The condition is $\Delta'=4a^2(1-t^2)=0$, hence $t=\pm 1$. The abscissae of the points of contact with the parabola are precisely the double roots, i.e. $x_1,x_2=2a$, so that $y_1,y_2=\pm 2a$.