What is the equation for a tangent to a tilted parabola?

Question

The equation of a parabola with directrix $ lx + my + n = 0 $ and focus $ P(x_P, y_P) $ is $$ \frac{(lx + my + n)^2}{l^2 + m^2} = (x - x_P)^2 + (y - y_P)^2 $$ What is the equation of a line tangent to this parabola?

I have tried many things like solving the equation for $ y $ and then differentiating with respect to $ x $ but I cannot manage to find an equation for a tangent.

Answer 1

You can differentiate the equation of the parabola implicitly (imagine that $y=y(x)$, and differentiate both sides of the equation normally). Solving $y'(x)=dy/dx$ from the resulting equation gives the following beauty $$ \frac{dy}{dx}=\frac{m^2x-\ell m y-\ell n-(m^2+\ell^2) x_P} {(\ell^2+m^2) y_P-\ell^2 y+\ell m x+m n}. $$

If you are given a point $P_0=(x_0,y_0)$ on your parabola, you can then plug in the values $x\leftarrow x_0$, $y\leftarrow y_0$ into this formula, and calculate the slope $k_0$ of the tangent at $P_0$: $$ k_0=\frac{m^2x_0-\ell m y_0-\ell n-(m^2+\ell^2) x_P} {(\ell^2+m^2) y_P-\ell^2 y_0+\ell m x_0+m n}. $$ The equation of the tangent is then the usual $$ y-y_0=k_0(x-x_0). $$ Warning: Because the parabola is tilted, it may happen that at some point the tangent will be vertical. You can diagnose this by seeing that the denominator in the formula for $k_0$ vanishes.

Answer 2

You can find the equation of the tangent at a point $A(x_A,\;y_A)$ which lies on the parabola with the following substitutions

$x^2\to x_Ax;\;y^2\to y_Ay;\;xy\to\dfrac{y_Ax+x_Ay}{2};\;x\to\dfrac{x+x_A}{2};\;y\to\dfrac{y+y_A}{2}$

For instance

Given the parabola $x^2+4 x y+4 y^2-4x-2 y=8$ the equation of the tangent at its point $Q(4;\;-4)$ is

$$4x +\frac{4(4 y-4 x)}{2} -4\cdot 4y-\frac{4 (x+4)}{2}-\frac{2 (y-4)}{2} =8$$

which gives $2 x+3 y+4=0$

Important notice: This formula works only if the point is a point of the curve and works for any conic

Answer 3

The polar line of a point on a conic is the tangent to the conic at that point.

The equation of this parabola can be rearranged into the form $$ m^2x^2-2lmxy+l^2y^2 -2\left(ln+(l^2+m^2)x_P\right)x-2\left(mn+(l^2+m^2)y_P\right)y+(l^2+m^2)(x_P^2+y_P^2)-n^2=0\tag{*}$$ or, in matrix form, $$\mathbf x^TC\mathbf x = \begin{bmatrix}x&y&1\end{bmatrix}\begin{bmatrix}m^2 & -lm & -ln-(l^2+m^2)x_P \\ -lm & l^2 & -mn-(l^2+m^2)y_P \\ -ln-(l^2+m^2)x_P & -mn-(l^2+m^2)y_P & (l^2+m^2)(x_P^2+y_P^2)-n^2\end{bmatrix}\begin{bmatrix}x\\y\\1\end{bmatrix} = 0.$$ The equation of the polar line to the point $\mathbf q=(x_q,y_q)$ is $\mathbf x^TC\mathbf q=0$, which I’ll leave to you to multiply out if you want a general solution. This amounts to making the substitutions described in Raffaele’s answer into equation (*).