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Suppose $f(x, y) =0$ represents a conic section, then $\sqrt{f(a, b)} =\sqrt{S11} $ ? is the length of tangents drawn to conic drawn from point $(a, b) $

This is easily done for circle but after that I have no idea where this is coming from and it's giving me an headache. It's given in my textbook without proof.

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  • $\begingroup$ Please use $MathJax$. Your equations will look like $x=0$ instead of the pathetic x=0. $\endgroup$ – Agile_Eagle Aug 12 '17 at 10:55
  • $\begingroup$ @Vrisk: Is my edit correct? And what is $S11$? $\endgroup$ – Emilio Novati Aug 12 '17 at 11:03
  • $\begingroup$ @EmilioNovati, the edit's alright. S11 is apparently a short hand for value of f(a, b). Does this look correct? $\endgroup$ – Vrisk Aug 12 '17 at 11:22
  • $\begingroup$ 1. If $f(x,y)=0$ represents a conic section, then $kf=0$ does also for any number $k$. 2. The two tangents have the same length segments to the points of intersection with a circle, this is not the case anymore for a general conic section. For $f=x^2-xy+y^2-1$ and tangents through $(-2,0)$ the segments are $\sqrt{2}$ and $\sqrt{5}$ and $\sqrt{f(-2,0)}=\sqrt{3}$. $\endgroup$ – Jan-Magnus Økland Aug 12 '17 at 14:22
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Given the conic equation $a_{11} x^2 + a_{22} y^2 + 2a_{12}xy + 2a_{13} x + 2a_{23} y + a_{33} = 0 $

The conic is non-degenerate if the determinant of the symmetric matrix $a_{ij}=a_{ji}$ is not zero

$$A=\left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{array} \right|\ne 0$$

you can write the equation of the polar line from an external point $P(x_0,y_0)$

$$(a_{11} x_0 + a_{12} y_0 + a_{13} )\,x + (a_{21} x_0 + a_{22} y_0 + a_{23} )\,y + a_{31} x_0 + a_{32} y_0 + a_{33} = 0$$

This line intersects the conic in the points where the tangents do, so you can solve the system between the conic and the polar and find the distances from the intersection points to the point $P$

In the example of Jan Magnus we had

$x^2-xy+y^2-1=0$ so that $a_{11}=1;\;a_{12}=a_{21}=-\frac12;\;a_{22}=1;\;a_{33}=-1$

and the point is $P(-2,0)$

Thus the polar line has equation

$-2x + \left(-\frac12 (-2) \right)\,y -1= 0 $ that is $y=2x+1$

plug this equation in the equation of the conic, to solve the system

$x^2-x(2x+1)+(2x+1)^2-1=0$ that is $3 x^2+3 x=0$ which gives $x_1=0;\;y_1=1$ and $x_2=-1;\;y_2=1$ Intersection points are $A(0,1)$ and $B(-1,1)$

We find $PA=\sqrt 5$ and $PB=\sqrt 2$

Hope this helps

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