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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.

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  • $\begingroup$ 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. $\endgroup$ – Jan-Magnus Økland Jan 8 at 13:35
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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 $\tan \theta = {2 \sqrt{12^2+9\cdot 4} \over 9-4}=\frac{12}{\sqrt{5}}.$

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