# Thread: Tangent lines to conics

1. ## Tangent lines to conics

I am unable to comprehend the proof for tangent line to conics. Here is the proof as per the book (Multiview Geometry by Hartley and Zisserman). Everything is in homogeneous coordinates.

The line l = Cx passes through x, since l(TRANSPOSE) x = x(TRANSPOSE) Cx = 0. If l has one-point contact with the conic, then it is a tangent, and we are done. Otherwise suppose that l meets the conic in another point y. Then y(TRANSPOSE) Cy = 0 and x(TRANSPOSE) Cy = l(TRANSPOSE) y = 0. From this it follows that (x + αy)(TRANSPOSE) C(x + αy) = 0 for all α, which means that the whole line l = Cx joining x and y lies on the conic C, which is therefore degenerate.

where C is conic coefficient matrix = [ a, b/2, d/2 ; b/2, c, e/2 ; d/2, e/2, f ]

I dont see how the underlined portion follows from the above premise. And even if it does how is the line a tangent to the conic?

2. Originally Posted by erohanip
I am unable to comprehend the proof for tangent line to conics. Here is the proof as per the book (Multiview Geometry by Hartley and Zisserman). Everything is in homogeneous coordinates.

The line l = Cx passes through x, since l(TRANSPOSE) x = x(TRANSPOSE) Cx = 0. If l has one-point contact with the conic, then it is a tangent, and we are done. Otherwise suppose that l meets the conic in another point y. Then y(TRANSPOSE) Cy = 0 and x(TRANSPOSE) Cy = l(TRANSPOSE) y = 0. From this it follows that (x + αy)(TRANSPOSE) C(x + αy) = 0 for all α, which means that the whole line l = Cx joining x and y lies on the conic C, which is therefore degenerate.

where C is conic coefficient matrix = [ a, b/2, d/2 ; b/2, c, e/2 ; d/2, e/2, f ]

I dont see how the underlined portion follows from the above premise. And even if it does how is the line a tangent to the conic?
You know that $x^{\textsc t}Cx = 0$ and $y^{\textsc t}Cy = 0$, because x and y lie on the conic. You are also assuming that $x^{\textsc t}Cy = 0$. Take the transpose of that to see that $y^{\textsc t}Cx = 0$. Therefore (multiplying out the brackets)

$(x+ay)^{\textsc t}C(x+ay) = x^{\textsc t}Cx + ax^{\textsc t}Cy + ay^{\textsc t}Cx + a^2y^{\textsc t}Cy = 0.$

What that equation says is that every point $x+ay$ on the line $l$ lies on the conic. So the whole line lies on the conic and is therefore tangent to it .