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Question: We have conic section $-12x^2 + 28xy+4x-9y^2-8y=0$ and a point not on a conic $(2/5,1/5)$, how to find an intersection point with tangent dropped from $(2/5,1/5)$ to a conic section?

My ideas: to use polar points and Homogeneous coordinates

Luidzzhi
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3 Answers3

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If you want to use pole-and-polar relation with homogeneous coordinates, matrix notation looks more elegant.

  • Conic

$$0= \begin{pmatrix} x & y & z \\ \end{pmatrix} \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

  • Pole

$$(x',y',z')$$

  • Polar

$$0= \begin{pmatrix} x' & y' & z' \\ \end{pmatrix} \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

  • In usual practice, we take $$z=z'=1$$

  • Polar is a tangent if the pole itself is on the conic; whereas if the pole "outside" the conic, the polar is the chord of contact. Intersection of the polar and the conic are your intersection.

See also another posts of mine here and here.

Ng Chung Tak
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  • I would be very grateful if you showed in more detail how to find the polar equation – Luidzzhi Apr 28 '22 at 08:08
  • See details for Joachimsthal's notation here. – Ng Chung Tak Apr 28 '22 at 10:12
  • If you are not familiar with matrix operation, please use the following fact: for a conic $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ the equation of polar (tangent/chord of contacts) at/from the pole $(x',y')$ is given by $$ax'x+h(y'x+x'y)+by'y+g(x+x')+f(y+y')+c=0$$ – Ng Chung Tak Apr 28 '22 at 10:21
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The given equation of the conic is

$ - 12 x^2 + 28 xy - 9 y^2 + 4 x - 8 y = 0 \hspace{15pt}(1) $

And the given point is $(\dfrac{2}{5}, \dfrac{1}{5})$

Using implicit differentiation, with respect to $x$,

$ - 24 x + 28 (y + x y') - 18 y y' + 4 - 8 y' = 0 $

Therefore,

$ y' (28 x - 18 y - 8) = 24 x - 28 y - 4 $

i.e.

$ y' = \dfrac{ 24 x - 28 y - 4 }{28 x - 18 y - 8} = \dfrac{12 x - 14 y - 2}{14 x - 9 y - 4 } $

Now $y'$ is the slope of the tangent line, which is also given by

$ y' = \text{Slope} = \dfrac{ \dfrac{1}{5} - y }{\dfrac{2}{5} - x} $

Therefore, by cross multiplication,

$ ( 12 x - 14 y - 2 ) (\dfrac{2}{5} - x) = (14 x - 9 y - 4) (\dfrac{1}{5} - y) $

Multiply through by $5$

$ (12 x - 14 y - 2) (2 - 5 x) = (14 x - 9 y - 4) ( 1 - 5 y)$

Expand

$ 24 x - 60 x^2 - 28 y+ 70 x y - 4 + 10 x = 14 x - 9 y - 4 - 70 x y + 45 y^2 + 20 y$

And this reduces to

$ -60 x^2 + 140 x y - 45 y^2 + 20 x -39 y = 0 \hspace{15pt} (2)$

Now we have to solve the system of equations $(1),(2)$, and the solutions are

$(0, 0) $ and $(\dfrac{1}{3} , 0 ) $

Hosam Hajeer
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Discriminant is a more straight forward way than polar and implicit differentiation. A conic (Eq.1) is tangent to a line (Eq.2) if and only if the following equations have only one real root pair: $$\begin{cases}ax^2+bxy+cy^2+dx+ey+f=0\enspace\enspace\text{Eq.1}\\gx+hy+k=0\enspace\enspace\text{Eq.2}\end{cases}$$ i.e. eliminating x or y and get a quadratic equation, the discriminant should be zero.

In this problem, we have: $$\begin{cases}−12x^2+28xy+4x−9y^2−8y=0\enspace\enspace\text{Eq.1}\\y-\dfrac{1}{5}=k(x-\dfrac{2}{5})\enspace\enspace\text{Eq.2}\end{cases}$$

By eliminating x we can get $$\frac{-9k^2+28k-12}{k^2}y^2+\frac{16k^2-56k+24}{5k^2}y+\frac{-8k^2+28k-12}{25k^2}=0$$ or by eliminating y we can get $$(-9k^2+28k-12)x^2+\frac{36k^2-114k+48}{5}x+\frac{-36k^2+116k-49}{25}=0$$ Either quadratic equation about k we choose, it should have one real root (i.e. double root). Let discriminant be zero ($b^2=4ac$) then we can solve that $$k_1=\frac{1}{2},k_2=3$$ So the tangent points ($y=-\dfrac{b}{2a}$ or $x=-\dfrac{b}{2a}$) are $(0,0)$ and $(\dfrac{1}{3},0)$.

auntyellow
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