# Angle between pair of tangents drawn from a point to a conic

The angle between the pair of tangents drawn to the ellipse 3x^2 + 2y^2 = 5 from the point (1,2) is?

I considered using homogenization for this problem, consider the shifted coordinates:

$$x' = x-1$$

$$y' = y-2$$

In shifted coordinates, our conic becomes:

$$3(x'+1)^2 + 2 (y'+2)^2 =5 \tag{1}$$

The relation of slope is given as:

$$3 (x'+1) + 2(y'+2) \frac{dy'}{dx} = 0 \tag{2}$$

Suppose the line passing the two intersection point is given as $$Ax+By =1$$, then homogenizing (1) with it,

$$3(x'+1)^2 + 2(y'+2)^2 = 5 (Ax'+By')^2 \tag{3}$$

This factorizes to the form:

$$(y'-m_1x ) ( y'-m_2 x) = 0 \tag{4}$$

Now, how do I solve for $$\{m_1,m_2 \}$$ from (3) and (4)? I'm not sure to how introduce (1) and (2) into helping me solve them.

Note: I've seen this question before already, but I want to solve it using homogenization.

• The line pair is $(3x^2 + 2y^2 - 5)(3\cdot 1^2 + 2\cdot 2^2 - 5)=(3x\cdot 1 + 2y\cdot 2 - 5)^2$ by Joachimsthal. – Jan-Magnus Økland Jan 8 at 13:35

## 1 Answer

If you have the line pair as $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ then the angle is given by $$\tan \theta = {2 \sqrt{h^2-ab} \over a+b}.$$

E.g. $$9x^2-24xy-4y^2+30x+40y-55=0$$ and $$\tan \theta = {2 \sqrt{12^2+9\cdot 4} \over 9-4}=\frac{12}{\sqrt{5}}.$$