Specific method for finding focus of oblique parabolas?

Question

Its easy to find focus of standard parabolas but how to find directrix,focus,vertex etc. of oblique parabolas(Whose axis is not parallel to x or y axis).Any help:)

Answer 1

If you could find the vertex and axis direction of the parabola, you could then proceed in several ways to compute all of those other properties of it.

The general conic equation $Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$ represents a parabola if the discriminant $B^2-AC=0$. This condition means that the quadratic part $Ax^2+2Bxy+Cy^2$ is a perfect square, which we can write as $(\alpha x+\beta y)^2$.

On the other hand, if the line $ax+by+c=0$ is a parabola’s axis and its tangent at the vertex is $bx-ax+d=0$, then an equation of the parabola is $$(ax+by+c)^2=4p(bx-ay+d).\tag{*}$$ If the coefficients of the two affine terms are normalized so that $a^2+b^2=1$, this is simply a rotated and translated version of the standard equation $y^2=4px$. I assume that you know how to extract the information that you’re interested in from that.

Comparing these two equations of the parabola, we see that we can find the direction of the parabola’s axis from the general conic equation by factoring the quadratic part. The vertex is the point at which the normal to the parabola is parallel to its axis. Writing the parabola’s equation as $f(x,y)=0$, this condition can be formulated as $$\det\begin{bmatrix} \beta & f_x(x,y) \\ -\alpha & f_y(x,y) \end{bmatrix} = \alpha f_x(x,y) + \beta f_y(x,y) = 0.$$ It turns out that this linear equation is in fact an equation of the parabola’s axis. With this equation in hand, you can rearrange the general conic equation into the form (*). The vertex of the parabola is the intersection of the axis and vertex tangent (or axis and parabola), which is easily found to be $$\left(-{ac+bd\over a^2+b^2},{ad-bc\over a^2+b^2}\right).$$

Alternatively, you can write the general equation in homogeneous matrix form $\mathbf x^TQ\mathbf x=0$. A parabola is tangent to the line at infinity and the point of tangency is the intersection of its axis and the line at infinity (i.e., the direction of the axis). This point is the pole of the line at infinity, which you can compute as $Q^{-1}(0,0,1)^T$, i.e., the last row or column of $Q^{-1}$, and will have the form $\mathbf p = (b,-a,0)^T$. The tangent at the vertex is then $\mathbf l = (b,-a, \tau)^T$ for some $\tau$, which you can find by solving the dual conic equation $\mathbf l^TQ^{-1}\mathbf l=0$ for $\tau$. The vertex is just the pole of this line, $\mathbf v = Q^{-1}\mathbf l$, and the axis is $\mathbf v\times\mathbf p$.